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Applications of Integration

2.4 Separable Differential Equations

37 problems · hints, answers and solutions shown beside each one

Recall that we are using log⁡x\log x to denote the logarithm of xx with base ee. In other courses it is often denoted ln⁡x\ln x.

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Below are pairs of functions y=f(x)y=f(x) and differential equations. For each pair, decide whether the function is a solution of the differential equation.

functiondifferential equation
(a)y=5(ex−3x2−6x−6)y=5(e^x-3x^2-6x-6)dydx=y+15x2\displaystyle \diff{y}{x}=y+15x^2
(b)y=−2x2+1y=\dfrac{-2}{x^2+1}y′(x)=xy2y'(x)=xy^2
(c)y=x3/2+xy=x^{3/2}+x(dydx)2+dydx=y\displaystyle\left(\diff{y}{x}\right)^2 + \diff{y}{x}=y
Hint

You don't need to solve the differential equation from scratch, only verify whether the given function y=f(x)y=f(x) makes it true. Find dydx\diff{y}{x} and plug it into the differential equation.

Answer

(a) yes (b) yes (c) no

Full solution

  1. If y=5(ex−3x2−6x−6)y=5(e^x-3x^2-6x-6), then dydx=5(ex−6x−6)\diff{y}{x} = 5(e^x-6x-6). Let's see whether this is equal to y+15x2y+15x^2:

    y+15x2=5(ex−3x2−6x−6)+15x2=5(ex−3x2−6x−6+3x2)=5(ex−6x−6)=dydx\begin{align*} y+15x^2&=5(e^x-3x^2-6x-6)+15x^2\\ &=5(e^x-3x^2-6x-6+3x^2)\\ &=5(e^x-6x-6)\\ &=\diff{y}{x} \end{align*}

    So, y=5(ex−3x2−6x−6)y=5(e^x-3x^2-6x-6) is indeed a solution to the differential equation dydx=y+15x2\diff{y}{x}=y+15x^2.

  2. If y=−2x2+1y=\dfrac{-2}{x^2+1}, then dydx=4x(x+1)2\diff{y}{x} = \dfrac{4x}{(x+1)^2}. Let's see whether this is equal to xy2xy^2:

    xy2=x(−2x2+1)2=4x(x2+1)2=dydx\begin{align*} xy^2&=x\left(\frac{-2}{x^2+1}\right)^2\\ &=\frac{4x}{(x^2+1)^2}\\ &=\diff{y}{x} \end{align*}

    So, y=−2x2+1y=\dfrac{-2}{x^2+1} is indeed a solution to the differential equation dydx=xy2\diff{y}{x}=xy^2.

  3. If y=x3/2+xy=x^{3/2}+x, then dydx=32x+1\diff{y}{x} = \frac{3}{2}\sqrt{x}+1.

    (dydx)2+dydx=(32x+1)2+32x+1=94x+92x+2≠dydx\begin{align*} \left(\diff{y}{x}\right)^2+\diff{y}{x}&=\left(\frac{3}{2}\sqrt{x}+1\right)^2+\frac{3}{2}\sqrt{x}+1\\ &=\frac{9}{4}x+\frac{9}{2}\sqrt{x}+2\\ &\neq \diff{y}{x} \end{align*}

    So, y=x3/2+xy=x^{3/2}+x is not a solution to the differential equation (dydx)2+dydx=y\left(\diff{y}{x}\right)^2+\diff{y}{x}=y.

Q2Stage 1

Following Definition 2.4.1 in the CLP-2 text, a separable differential equation has the form

dydx(x)=f(x) g(y(x)).\diff{y}{x}(x) = f(x)\ g\big(y(x)\big).

Show that each of the following equations can be written in this form, identifying f(x)f(x) and g(y)g(y).

  1. 3ydydx=xsin⁡y3y\diff{y}{x}=x\sin y

  2. dydx=ex+y\diff{y}{x} = e^{x+y}

  3. dydx+1=x\diff{y}{x}+1=x

  4. (dydx)2−2xdydx+x2=0\left(\diff{y}{x}\right)^2-2x\diff{y}{x}+x^2=0

Hint

For (d), note the equation given is quadratic in the variable dydx\diff{y}{x}.

Answer

  1. One possible answer: f(x)=xf(x)=x, g(y)=sin⁡y3yg(y)=\dfrac{\sin y}{3y}.

  2. One possible answer: f(x)=exf(x) = e^x, g(y)=eyg(y) = e^y.

  3. One possible answer: f(x)=x−1f(x) = x-1, g(y)=1g(y) = 1.

  4. The given equation is equivalent to the equation dydx=x\diff{y}{x}=x, which fits the form of a separable equation with f(x)=xf(x)=x, g(y)=1g(y)=1.

Full solution

  1. 3ydydx=xsin⁡y3y\diff{y}{x}=x\sin y can be written as dydx=x(sin⁡y3y)\diff{y}{x} = x\left(\frac{\sin y}{3y}\right), which fits the form of a separable equation with f(x)=xf(x)=x, g(y)=sin⁡y3yg(y) = \frac{\sin y}{3y}.

  2. dydx=ex+y=exey\diff{y}{x} = e^{x+y} = e^xe^y which fits the form of a separable equation using f(x)=exf(x) = e^x, g(y)=eyg(y) = e^y.

  3. dydx+1=x\diff{y}{x}+1=x can be written as dydx=(x−1)\diff{y}{x} = (x-1), which fits the form of a separable equation using f(x)=x−1f(x)=x-1, g(y)=1g(y)=1. (We can solve it by simply antidifferentiating.)

  4. Notice the left side of the equation (dydx)2−2xdydx+x2=0\left(\diff{y}{x}\right)^2-2x\diff{y}{x}+x^2=0 is a perfect square. So, this equation is equivalent to (dydx−x)2=0\left(\diff{y}{x}-x\right)^2=0, that is, dydx=x\diff{y}{x}=x. This has the form of a separable equation with f(x)=xf(x)=x, g(y)=1g(y)=1.

Q3Stage 1

Suppose we have the following functions:

  • yy is a differentiable function of xx

  • ff is a function of xx, with ∫f(x) dx=F(x)\int f(x)\,\dee{x}=F(x)

  • gg is a nonzero function of yy, with ∫1g(y) dy=G(y)=G(y(x))\int \frac{1}{g(y)} \,\dee{y}=G(y)=G(y(x)).

In the work below, we set up a solution to the separable differential equation

dydx=f(x) g(y)=f(x) g(y(x))\diff{y}{x}=f(x)\,g(y)=f(x)\,g(y(x))

without using the mnemonic of Equation 2.4.1 in the CLP-2 text.

By deleting some portion of our work, we can create the solution as it would look using the mnemonic. What portion can be deleted?

Remark: the purpose of this exercise is to illuminate what, exactly, the mnemonic is a shortcut for. Despite its peculiar look, it agrees with what we already know about integration.

dydx=f(x)g(y(x))\begin{align*}\diff{y}{x}&=f(x)g(y(x))\end{align*}

Since g(y(x))g(y(x)) is a nonzero function, we can divide both sides by it.

1g(y(x))⋅dydx=f(x)\begin{align*}\frac{1}{g(y(x))}\cdot\diff{y}{x}&=f(x)\end{align*}

If these functions of xx are the same, then they have the same antiderivative with respect to xx.

∫1g(y(x))⋅dydx dx=∫f(x) dx\begin{align*}\int \frac{1}{g(y(x))}\cdot\diff{y}{x}\,\dee{x}&=\int f(x)\,\dee{x}\end{align*}

The left integral is in the correct form for a change of variables to yy. To make this easier to see, we'll use a uu-substitution, since it's a little more familiar than a yy-substitution. If u=yu=y, then dudx=dydx\diff{u}{x}=\diff{y}{x}, so du=dydxdx\dee{u}=\diff{y}{x}\dee{x}.

∫1g(u) du=∫f(x) dx\begin{align*}\int\frac{1}{g(u)}\,\dee{u}&=\int f(x)\,\dee{x}\end{align*}

Since uu was just the same as yy, again for cosmetic reasons, we can swap it back. (Formally, you could have skipped the step above–we just included it to be extra clear that we're not using any integration techniques we haven't seen before.)

∫1g(y) dy=∫f(x) dx\begin{align*}\int\frac{1}{g(y)}\,\dee{y}&=\int f(x)\,\dee{x}\end{align*}

We're given the antiderivatives in question.

G(y)+C1=F(x)+C2G(y)=F(x)+(C2−C1)\begin{align*}G(y)+C_1&=F(x)+C_2\\ G(y)&=F(x)+(C_2-C_1)\end{align*}

where C1C_1 and C2C_2 are arbitrary constants. Then also C2−C1C_2-C_1 is an arbitrary constant, so we might as well call it CC.

G(y)=F(x)+C\begin{align*}G(y)&=F(x)+C\end{align*}
Hint

The step ∫1g(y) dy=∫f(x) dx\displaystyle\int \frac{1}{g(y)}\,\dee{y} = \int f(x)\,\dee{x} shows up whether we're using our mnemonic or not.

Answer

The mnemonic allows us to skip from the separable differential equation we want to solve (very first line) to the equation

∫1g(y) dy=∫f(x) dx\int \frac{1}{g(y)}\,\dee{y}=\int f(x)\,\dee{x}
Full solution

The mnemonic allows us to skip from the separable differential equation we want to solve (very first line) to the equation

∫1g(y) dy=∫f(x) dx\int \frac{1}{g(y)}\,\dee{y}=\int f(x)\,\dee{x}

So, the mnemonic is just a shortcut for the substitution we performed to get this point.

We also generally skip the explanation about C1C_1 and C2C_2 being replaced with CC.

Q4Stage 1

Suppose y=f(x)y=f(x) is a solution to the differential equation dydx=xy\diff{y}{x}=xy.

True or false: f(x)+Cf(x)+C is also a solution, for any constant CC.

Hint

Note ddx{f(x)}=ddx{f(x)+C}\diff{}{x}\{f(x)\} = \diff{}{x}\{f(x)+C\}. Plug in y=f(x)+Cy=f(x)+C to the equation dydx=xy\diff{y}{x}=xy to see whether it makes the equation is true.

Answer

false

Full solution

To say y=f(x)+Cy=f(x)+C is a solution to the differential equation means:

ddx{f(x)+C}=x(f(x)+C)\begin{align*}\diff{}{x}\{f(x)+C\}&=x(f(x)+C)\end{align*}

Since y=f(x)y=f(x) is a solution, we know ddx{f(x)}=xf(x)\diff{}{x}\{f(x)\}=xf(x). Also, ddx{f(x)+C}=ddx{f(x)}\diff{}{x}\{f(x)+C\}=\diff{}{x}\{f(x)\}. So, ddx{f(x)+C}=xf(x)\diff{}{x}\{f(x)+C\}=xf(x).

xf(x)=x(f(x)+C)0=xC\begin{align*}xf(x)&=x(f(x)+C)\\ 0&=xC\end{align*}

Our equation should hold for all xx in our domain, and for the derivative to yy with respect to xx to make sense, our domain should not be a single point. So, there is some xx in our domain such that x≠0x \neq 0. Therefore, the CC must be zero. So, f(x)+Cf(x)+C is not a solution to the differential equation for any constant CC.

When we're finding a general antiderivative, we add “+C+C" at the end. When we're finding a general solution to a differential equation, the “+C+C" gets added when we antidifferentiate–we don't add another one at the end of our work.

Q5Stage 1

Suppose a function y=f(x)y=f(x) satisfies ∣y∣=Cx|y| = Cx, for some constant C>0C>0.

  1. What is the largest possible domain of f(x)f(x), given the information at hand?

  2. Give an example of function y=f(x)y=f(x) with the following properties, or show that none exists:

    • ∣y∣=Cx|y| = Cx,

    • dydx\diff{y}{x} exists for all x>0x>0, and

    • y>0y>0 for some values of xx, and y<0y<0 for others.

Hint

If a function is differentiable at a point, it is also continuous at that point.

Answer

(a) [0,∞)[0,\infty)
(b) No such function exists. If ∣f(x)∣=Cx|f(x)|=Cx and f(x)f(x) switches from f(x)=Cxf(x)=Cx to f(x)=−Cxf(x)=-Cx at some point, then that point is a jump discontinuity. Where f(x)f(x) contains a discontinuity, dydx\diff{y}{x} does not exist.

Full solution

  1. Since ∣y∣≥0|y| \geq 0 no matter what yy is, we see Cx≥0Cx \ge 0 for all xx in the domain of f(x)f(x). Since CC is positive, that means the domain of f(x)f(x) only includes nonnegative numbers. So, the largest possible domain of f(x)f(x) is [0,∞)[0,\infty).

  2. None exists.

    The graph of CxCx is given below for some positive constant CC, also with the graph of −Cx-Cx. If y=f(x)y=f(x) were sometimes the top function, and other times the bottom function, then there would be a jump discontinuity where it switched. Then the derivative of f(x)f(x) would not exist, violating the second property.

    Figure from prob_s2.4, line 1

    Figure from prob_s2.4, line 1

    A tiny technical note is that it's possible that f(x)=Cxf(x)=Cx when x=0x=0 and f(x)=−Cxf(x)=-Cx when x>0x>0 (or vice-versa). This would not introduce a jump discontinuity, but it also does not satisfy that f(x)>0f(x)>0 for some values of xx.

Remark: in several instances below, solving a differential equation will lead us to conclude something like ∣y∣=g(x)|y|=g(x). In these cases, we choose either y=g(x)y=g(x), or y=−g(x)y=-g(x), but not y=±g(x)y=\pm g(x) (which is not a function) or that yy is sometimes g(x)g(x), and other times −g(x)-g(x). The reasoning above somewhat explains this choice: if yy were sometimes positive and sometimes negative, then dydx\diff{y}{x} would not exist at the values of xx where the sign of yy switches, unless that switch occurrs at a root of g(x)g(x). Since that's a pretty specific occurrence, we usually feel safe ignoring it to avoid getting bogged down in technical details.

Q6Stage 1

Express the following sentence (The sentence is paraphrased from the Pharmakokinetics website of Universit'e de Lausanne, “Elimination Kinetics," at https://sepia.unil.ch/pharmacology/index.php?id=94 . The half-life of morphine is given on the same website at https://sepia.unil.ch/pharmacology/index.php?id=85 . Accessed 12 August 2017.) as a differential equation. You don't have to solve the equation.

About 0.3 percent of the total quantity of morphine in the bloodstream is eliminated every minute.

Hint

Let Q(t)Q(t) be the quantity of morphine in a patient's bloodstream at time tt, where tt is measured in minutes.

Using the definition of a derivative,

dQdt=lim⁡h→0Q(t+h)−Q(t)h≈Q(t+1)−Q(t)1\diff{Q}{t}=\lim_{h \to 0}\frac{Q(t+h)-Q(t)}{h}\approx \frac{Q(t+1)-Q(t)}{1}

So, dQdt\diff{Q}{t} is roughly the change in the amount of morphine in one minute, from tt to t+1t+1.

Answer

dQdt=−0.003Q(t)\displaystyle \diff{Q}{t}=-0.003Q(t)

Full solution

Let Q(t)Q(t) be the quantity of morphine in a patient's bloodstream at time tt, where tt is measured in minutes.

Using the definition of a derivative,

dQdt=lim⁡h→0Q(t+h)−Q(t)h≈Q(t+1)−Q(t)1\diff{Q}{t}=\lim_{h \to 0}\frac{Q(t+h)-Q(t)}{h}\approx \frac{Q(t+1)-Q(t)}{1}

So, dQdt\diff{Q}{t} is roughly the change in the amount of morphine in one minute, from tt to t+1t+1.

The sentence tells us that the change in the amount of morphine in one minute is about −0.003Q-0.003Q, where QQ is the quantity in the bloodstream. That is:

dQdt=−0.003Q(t)\diff{Q}{t}=-0.003Q(t)
Q7Stage 1

Suppose a particular change is occurring in a language, from an old form to a new form. (An example is the change in German from “wollt" to “wollst" for the second-person conjugation of the verb “wollen." This example is provided by the site Laws in Quantitative Linguistics, “Change in Language" http://lql.uni-trier.de/index.php/Change_in_language accessed 18 August 2017.) Let p(t)p(t) be the proportion (measured as a number between 0, meaning none, and 1, meaning all) of the time that speakers use the new form. Piotrowski's law (Piotrowski's law is paraphrased from the page Piotrowski-Gesetz on Glottopedia, http://www.glottopedia.org/index.php/Piotrowski-Gesetz, accessed 18 August 2017. According to this source, the law was based on work by the married couple R. G. Piotrowski and A. A. Piotrowskaja, later generalized by G. Altmann.) predicts the following.

Use of the new form over time spreads at a rate that is proportional to the product of the proportion of the new form and the proportion of the old form.

Express this as a differential equation. You do not need to solve the differential equation.

Hint

If p(t)p(t) is the proportion of the new form, then 1−p(t)1-p(t) is the proportion of the old form.

When we say two quantities are proportional, we mean that one is a constant multiple of the other.

Answer

dpdt=αp(t)(1−p(t))\diff{p}{t}=\alpha p(t)\big(1-p(t)\big), for some constant α\alpha.

Full solution

If p(t)p(t) is the proportion of times speakers use the new form, measured between 0 and 1, then 1−p(t)1-p(t) is the proportion of times speakers use the old form.

The law, then, states that dpdt\diff{p}{t} is proportional to p(t)×(1−p(t))p(t)\times \big(1-p(t)\big). When we say two quantities are proportional, we mean that one is a constant multiple of the other. So, the law says

dpdt=αp(t)(1−p(t))\diff{p}{t}=\alpha p(t)\big(1-p(t)\big)

for some constant α\alpha.

Remark: it follows from this model that, when a new form is either very rare or entirely ubiquitous, the rate of change of its adoption is small. This makes sense: if the new form is used all the time (p(t)≈1p(t)\approx 1), there's nobody left to convert; if the new form is almost never used (p(t)≈0p(t)\approx 0) then people don't know about it, so they won't pick it up.

Q8Stage 1

Consider the differential equation y′=y2−1y'=\frac{y}{2}-1.

  1. When y=0y=0, what is y′y'?

  2. When y=2y=2, what is y′y'?

  3. When y=3y=3, what is y′y'?

  4. On the axes below, interpret the marks we have made, and use them to sketch a possible solution to the differential equation.

Figure from prob_s2.4, line 2

Figure from prob_s2.4, line 2

Hint

The red marks show the slope y(x)y(x) would have at a point if it crosses that point. So, pick a value of y(0)y(0); based on the red marks, you can see how fast y(x)y(x) is increasing or decreasing at that point, which leads you roughly to a value of y(1)y(1); again, the red marks tell you how fast y(x)y(x) is increasing or decreasing, which leads you to a value of y(2)y(2), etc (unless you're already off the graph).

Answer

(a) −1-1 (b) 00 (c) 0.50.5
(d) Two possible answers are shown below:

Figure from prob_s2.4, line 2

Figure from prob_s2.4, line 2

Figure from prob_s2.4, line 14

Figure from prob_s2.4, line 14

Another possible answer is the constant function y=2y=2.

Full solution

  1. When y=0y=0, y′=02−1=−1y'=\frac{0}{2}-1=-1.

  2. When y=2y=2, y′=22−1=0y'=\frac{2}{2}-1=0.

  3. When y=3y=3, y′=32−1=0.5y'=\frac{3}{2}-1=0.5.

  4. The small red lines have varying slopes. The red lines on points with yy-coordinate 2 have slopes of 00; this matches y′y' when y=0y=0, as we saw above. The red lines on points with yy-coordinate 0 have slopes of approximately −1-1; again, this matches what we found for y′y' when y=0y=0.

    The red lines correspond to a tiny section of y(x)y(x), if y(x)y(x) passes through that point. So, we can sketch a possible curve y(x)y(x) satisfying the equation by starting somewhere, then following the slopes.

    For example, suppose we start at the origin.

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

    Then our function is decreasing at that point, which leads us to a coordinate where (as we see from the red marks) the function is decreasing slightly faster.

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

    Following the red marks leads us down even further, so our function y(x)y(x) might look something like this:

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

    However, we didn't have to start at the origin. Suppose y(0)=3y(0)=3. Then at x=0x=0, yy is increasing, with slope 12\frac{1}{2}.

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

    Our red marks run out that high up, but we now y′=12y−1y'=\frac{1}{2}y-1, so y′y' increases as yy increases. That means our function keeps getting steeper and steeper, possibly something like this:

    Figure from prob_s2.4, line 14

    Figure from prob_s2.4, line 14

    If y(0)=2y(0)=2, we see another possible curve is the constant function y(x)=2y(x)=2.

Remark: from Theorem 2.4.4 in the CLP-2 text, we see the solutions to the equation y′=12y−1=12(y−2)y'=\frac{1}{2}y-1 = \frac{1}{2}(y-2) are of the form y(x)=Cex/2+2y(x)=Ce^{x/2}+2 for some constant CC. Check that the curves you're sketching look exponential.

Q9Stage 1

Consider the differential equation y′=y−x2y'=y-\frac{x}{2}.

  1. If y(1)=0y(1)=0, what is y′(1)y'(1)?

  2. If y(1)=2y(1)=2, what is y′(1)y'(1)?

  3. If y(1)=−2y(1)=-2, what is y′(1)y'(1)?

  4. Draw a sketch similar to that of Question 8(d) showing the derivatives of yy at the points with integer values for xx in [0,6][0,6] and yy in [−3,3][-3,3].

  5. Sketch a possible graph of yy.

Hint

To draw the sketch similar to Question 8(d), don't actually calculate every single slope; find a few (for instance, where the slope is zero, or where it's negative), and use a pattern (for instance, the slope increases as yy increases) to approximate most of the points.

Answer

(a) −12-\dfrac{1}{2} (b) 32\dfrac{3}{2} (c) −52-\dfrac{5}{2}
(d) Your sketch should look something like this:

Figure from prob_s2.4, line 2

Figure from prob_s2.4, line 2

(e) There are lots of possible answers. Several are shown below.

Figure from prob_s2.4, line 2

Figure from prob_s2.4, line 2

Figure from prob_s2.4, line 16

Figure from prob_s2.4, line 16

Figure from prob_s2.4, line 31

Figure from prob_s2.4, line 31

Figure from prob_s2.4, line 46

Figure from prob_s2.4, line 46

Full solution

  1. If y(1)=0y(1)=0, then y′(1)=0−12=−12y'(1)=0-\frac{1}{2}=-\frac{1}{2}.

  2. If y(1)=2y(1)=2, then y′(1)=2−12=32y'(1)=2-\frac{1}{2}=\frac{3}{2}.

  3. If y(1)=−2y(1)=-2, then y′(1)=−2−12=−52y'(1)=-2-\frac{1}{2}=-\frac{5}{2}.

  4. There are 7×7=497 \times 7 = 49 points on the grid; we don't want to make 49 separate calculations. Let's find some shortcuts.

    • If y′(x)=0y'(x)=0, then y=x2y=\frac{x}{2}, which applies to the points (0,0)(0,0), (2,1)(2,1), (4,2)(4,2) and (6,3)(6,3). These are the orange dots in the sketch below.

    • If y′(x)=1y'(x)=1, then y=1+x2y=1+\frac{x}{2}, which applies to the points (0,1)(0,1), (2,2)(2,2), and (4,3)(4,3). (Note these are exactly 1 unit above the points with y′=0y'=0.) These are the red dots in the sketch below.

    • If y′(x)=−1y'(x)=-1, then y=−1+x2y=-1+\frac{x}{2}, which applies to the points (0,−1)(0,-1), (2,−2)(2,-2), and (4,−3)(4,-3). (Note these are exactly 1 unit below the points with y′=0y'=0.) These are the yellow dots in the sketch below.

    • If xx increases and yy stays the same, yy decreases.

    • If yy increases and xx stays the same, yy increases.

    • If we draw a straight line of slope 12\dfrac{1}{2} on our sketch, for every point on that line, our mark has the same slope: for instance, the points where we draw a mark with slope 0 are (0,0)(0,0), (2,1)(2,1), and (4,2)(4,2), and these all lie on the line f(x)=x2f(x)=\frac{x}{2}.

    This is enough to give us a pretty good sketch. The points whose slopes we found explicitly have dots; the rest can be sketched as either steeper or less steep than what's near them.

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

  5. To sketch a possible graph of y(x)y(x), we choose a point (x,y(x))(x,y(x)), then follow the red lines.

    For example, if we suppose that y(4)=2y(4)=2, then near (4,2),(4,2), the lines tell us y(x)y(x) is fairly flat; and it is increasing to the left of x=4x=4, and decreasing to the right.

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

    Following the red lines a little farther in each direction brings us somewhere like this:

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

    Extending yet further, we might sketch something like the following:

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

    By choosing another point (x,y(x))(x,y(x)) to be on the curve, we might find other potential curves. Some examples are shown below.

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 2

    Figure from prob_s2.4, line 16

    Figure from prob_s2.4, line 16

    Figure from prob_s2.4, line 31

    Figure from prob_s2.4, line 31

    Figure from prob_s2.4, line 46

    Figure from prob_s2.4, line 46

Remark: the differential equation y′=y−x2y'=y-\frac{x}{2} is not separable, so we haven't talked about how to solve it. The solutions have the form y(x)=Cex+x+12y(x)=Ce^{x}+\frac{x+1}{2}. You can verify that these functions satisfy y′=y−x2y'=y-\frac{x}{2}.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q10Stage 2Past exam · 2016Q5

Find the solution to the separable initial value problem:

dydx=2xey,y(0)=log⁡2\begin{align*} \diff{y}{x} &= \frac{2x}{e^y}, &y(0) &= \log 2 \end{align*}

Express your solution explicitly as y=y(x)y = y(x).

Hint

Start by multiplying both sides of the equation by eye^y and dx\dee{x}, pretending that dydx\diff{y}{x} is a fraction, according to our mnemonic.

Answer

y=log⁡(x2+2)y = \log(x^2+2)

Full solution

Rearranging, we have:

ey dy=2x dx.\begin{align*}e^y\,\dee{y} &= 2x\,\dee{x}.\end{align*}

Integrating both sides:

∫ey dy=∫2x dxey=x2+C\begin{align*}\int e^y\,\dee{y} &= \int 2x\,\dee{x}\\ e^y &= x^2+C\end{align*}

Since y=log⁡2y=\log2 when x=0x=0, we have

elog⁡2=02+C2=C,\begin{align*}e^{\log 2} &= 0^2+C \\ 2 &= C,\end{align*}

and therefore

ey=x2+2y=log⁡(x2+2)\begin{align*}e^{y} &= x^2+2 \\ y &= \log(x^2+2)\end{align*}
Q11Stage 2Past exam · M121 2002A

Find the solution y(x)y(x) of dydx=xyx2+1\displaystyle\diff{y}{x}=\frac{xy}{x^2+1}, y(0)=3y(0)=3.

Hint

You need to solve for your function y(x)y(x) explicitly. Be careful with absolute values: if ∣y∣=F|y|=F, then y=Fy=F or y=−Fy=-F. However, y=±Fy=\pm F is not a function. You have to choose one: y=Fy=F or y=−Fy=-F.

Answer

y(x)=31+x2y(x)=3\sqrt{1+x^2}

Full solution

Using separation of variables:

dydx=xyx2+1dyy=xx2+1dx∫dyy=∫xx2+1dxlog⁡∣y∣=12log⁡(1+x2)+C\begin{align*}\diff{y}{x}&=\frac{xy}{x^2+1}\\ \frac{\dee{y}}{y}&=\frac{x}{x^2+1}\dee{x}\\ \int\frac{\dee{y}}{y}&=\int\frac{x}{x^2+1}\dee{x}\\ \log |y| &= \frac{1}{2}\log(1+x^2)+C\end{align*}

To satisfy y(0)=3y(0)=3, we need log⁡3=12log⁡(1+0)+C\log 3 = \frac{1}{2}\log(1+0)+C, so C=log⁡3C=\log 3. Thus:

log⁡∣y∣=12log⁡(1+x2)+log⁡3=log⁡1+x2+log⁡3=log⁡31+x2\begin{align*}\log |y| &= \frac{1}{2}\log(1+x^2)+\log 3\\ &=\log\sqrt{1+x^2}+\log 3\\&=\log 3\sqrt{1+x^2}\end{align*}

So,

∣y∣=31+x2\begin{align*}|y|&=3\sqrt{1+x^2}\end{align*}

We are told to find a function y(x)y(x). So far, we have two possible functions from the work above: maybe y=31+x2y=3\sqrt{1+x^2}, and maybe y=−31+x2y=-3\sqrt{1+x^2}. It's important to note that y=±31+x2y=\pm3\sqrt{1+x^2} is not a function: for an equation to represent a function, for every input in the domain, there must only be one output. That is, functions pass the vertical line test. (Definition 0.4.1 in the CLP-1 text gives a formal definition of a function.) So, we need to decide whether our function is y=31+x2y=3\sqrt{1+x^2} or y=−31+x2y=-3\sqrt{1+x^2}. Since y(0)=3y(0)=3, we conclude

y(x)=31+x2\begin{align*}y(x)&=3\sqrt{1+x^2}\end{align*}
Q12Stage 2Past exam · M105 2015A

Solve the differential equation y′(t)=ey3cos⁡ty'(t)=e^{\frac{y}{3}}\cos t. You should express the solution y(t)y(t) in terms of tt explicitly.

Hint

If your answer doesn't quite look like the answer given, try manipulating it with logarithm rules: log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log(ab), and alog⁡b=log⁡(ba)a\log b = \log(b^a).

Answer

y(t)=3log⁡(−3C+sin⁡t)y(t)=3\log\left(\dfrac{-3}{C+\sin t}\right)

Full solution

The given differential equation is separable and we solve it accordingly.

y′=ey3cos⁡te−y/3dy=cos⁡t dt∫e−y/3dy=∫cos⁡t dt−3e−y/3=sin⁡t+C1ey/3=sin⁡t+C−3ey/3=−3C+sin⁡ty3=log⁡(−3C+sin⁡t)y(t)=3log⁡(−3C+sin⁡t)\begin{align*} y'&=e^{\frac{y}{3}}\cos t\\ e^{-y/3}\dee{y}&=\cos t\,\dee{t}\\ \int e^{-y/3}\dee{y}&=\int\cos t\,\dee{t}\\ -3e^{-y/3}&=\sin t +C\\ \frac{1}{e^{y/3}}&=\frac{\sin t +C}{-3}\\ e^{y/3} &=\frac{-3}{C+\sin t} \\ \frac{y}{3}&=\log\left(\frac{-3}{C+\sin t}\right) \\y(t)&=3\log\left(\frac{-3}{C+\sin t}\right) \end{align*}

for any constant CC.

Since the domain of logarithm is (0,∞)(0,\infty), the solution only exists when C+sin⁡t<0C+\sin t<0.

Q13Stage 2Past exam · 2000D

Solve the differential equation

dydx=xex2−log⁡(y2)\begin{equation*} \diff{y}{x} = x e^{x^2-\log(y^2)} \end{equation*}
Hint

Simplify the equation.

Answer

y=32ex2+C3y=\sqrt[3]{\frac{3}{2} e^{x^2}+C}.

Full solution

The given differential equation is separable and we solve it accordingly.

dydx=xex2−log⁡(y2)=xex2y2y2dy=xex2 dx∫y2dy=∫xex2 dx\begin{align*}\diff{y}{x} &= x e^{x^2-\log(y^2)} =\frac{x e^{x^2}}{y^2}\\ y^2\dee{y}&=x e^{x^2}\,\dee{x}\\ \int y^2\dee{y}&=\int x e^{x^2}\,\dee{x}\end{align*}

We can guess the antiderivative of xex2xe^{x^2}, or use the substitution u=x2u=x^2, du=2xdx\dee{u}=2x\dee{x}.

y33=12ex2+C′y3=32ex2+3C′\begin{align*}\frac{y^3}{3}&=\frac{1}{2} e^{x^2}+C'\\ y^3&=\frac{3}{2} e^{x^2}+3C'\end{align*}

Since C′C' can be any constant in (−∞,∞)(-\infty,\infty), then also 3C′3C' can be any constant in (−∞,∞)(-\infty,\infty), so we replace 3C′3C' with the arbitrary constant CC.

y3=32ex2+Cy=32ex2+C3\begin{align*}y^3&=\frac{3}{2} e^{x^2}+C \\y&=\sqrt[3]{\frac{3}{2} e^{x^2}+C}\end{align*}

for any constant CC.

Q14Stage 2Past exam · 2000D

Let y=y(x)y=y(x). Find the general solution of the differential equation y′=xeyy'=xe^y.

Hint

Be careful with the arbitrary constant.

Answer

y=−log⁡(C−x22)\displaystyle y=-\log\left(C-\frac{x^2}{2}\right)
The solution only exists for C−x22>0C-\frac{x^2}{2}>0, i.e. C>0C>0 and the function has domain {x:∣x∣<2C}\left\{x:|x|<\sqrt{2C}\right\}.

Full solution

The given differential equation is separable and we solve it accordingly.

dydx=xeydyey=x dx∫dyey=∫x dx−e−y=12x2+Ce−y=−12x2−C\begin{align*}\diff{y}{x}&=xe^y\\ \frac{\dee{y}}{e^y}&=x\,\dee{x}\\ \int\frac{\dee{y}}{e^y}&=\int x\,\dee{x}\\ -e^{-y}&=\frac{1}{2} x^2+C\\ e^{-y}&=-\frac{1}{2} x^2-C\end{align*}

Since CC can be any constant in (−∞,∞)(-\infty,\infty), then also −C-C can be any constant in (−∞,∞)(-\infty,\infty), so we write CC instead of −C-C.

e−y=C−12x2−y=log⁡(C−x22)y=−log⁡(C−x22)\begin{align*}e^{-y}&=C-\frac{1}{2} x^2 \\-y&=\log\left(C-\frac{x^2}{2}\right) \\y&=-\log\left(C-\frac{x^2}{2}\right)\end{align*}

for any constant CC.

The solution only exists for C−x22>0C-\frac{x^2}{2}>0. For this to happen, we need C>0C>0, and then the domain of the function is those values xx for which ∣x∣<2C|x|<\sqrt{2C}.

Q15Stage 2Past exam · 2016Q5

Find the solution to the differential equation yy′ex−2x=1y\displaystyle\frac{y y'}{e^x -2x} = \frac{1}{y} that satisfies y(0)=3y(0) = 3. Solve completely for yy as a function of xx.

Hint

Start by cross-multiplying.

Answer

y=(3ex−3x2+24)1/3y = (3e^x -3x^2+ 24)^{1/3}

Full solution

The given differential equation is separable and we solve it accordingly. Cross–multiplying, we rewrite the equation as

y2dydx=ex−2xy2 dy=(ex−2x) dx.\begin{align*}y^2 \diff{y}{x} &= e^x -2x\\ y^2 \,\dee{y} &= (e^x - 2x ) \,\dee{x}.\end{align*}

Integrating both sides, we find

∫y2 dy=∫(ex−2x) dx13y3=ex−x2+C\begin{align*}\int y^2 \,\dee{y} &=\int (e^x - 2x ) \,\dee{x}\\ \frac{1}{3}y^3 &= e^x - x^2 + C\end{align*}

Setting x=0x = 0 and y=3y = 3, we find 1333=e0−02+C\frac{1}{3}3^3=e^0-0^2+C and hence C=8C=8.

13y3=ex−x2+8y=(3ex−3x2+24)1/3\begin{align*}\frac{1}{3}y^3 &= e^x - x^2 + 8\\ y &= (3e^x -3x^2+ 24)^{1/3}\end{align*}
Q16Stage 2Past exam · 2016Q5

Find the function y=f(x)y=f(x) that satisfies

dydx=−xy3andf(0)=−14\begin{align*} \diff{y}{x} = -xy^3 \qquad\text{and}\qquad f(0)=-\frac{1}{4} \end{align*}
Hint

Be careful about signs. If y2=Fy^2=F, then possibly y=Fy=\sqrt{F}, and possibly y=−Fy=-\sqrt{F}. However, y=±Fy=\pm\sqrt{F} is not a function.

Answer

y=f(x)=−1x2+16y=f(x) = -\dfrac{1}{\sqrt{x^2+16}}

Full solution

This is a separable differential equation that we solve in the usual way.

dydx=−xy3−dyy3=x dx∫−dyy3=∫x dx−y−2−2=x22+Cy−2=x2+2C.\begin{align*}\diff{y}{x} &= -xy^3\\ - \frac{\dee{y}}{y^3} &= x\ \dee{x}\\ \int -\frac{\dee{y}}{y^3} &= \int x\ \dee{x}\\ - \frac{y^{-2}}{-2} &= \frac{x^2}{2}+C\\ y^{-2}&=x^2+2C. \tag{$*$}\end{align*}

To have y=−14y=-\frac{1}{4} when x=0x=0, we must choose CC to obey

(−14)−2=0+2C16=2C\begin{align*}{\Big(-\frac{1}{4} \Big)}^{-2}& =0 +2C\\ 16&= 2C\end{align*}

So, from (∗*),

y−2=x2+2C=x2+16y2=1x2+16\begin{align*}y^{-2}&=x^2+2C =x^2+16\\ y^2&=\frac{1}{x^2+16}\end{align*}

Now, we have two potential candidates for y(x)y(x):

y=1x2+16ORy=−1x2+16\begin{align*}y&=\frac{1}{\sqrt{x^2+16}}\qquad\text{OR}\qquad y=-\frac{1}{\sqrt{x^2+16}}\end{align*}

We know y=−14y=-\frac14 when x=0x=0. The only function above that fits this is

y=−1x2+16\begin{align*}y&=-\frac{1}{\sqrt{x^2+16}}\end{align*}

So, f(x)=−1x2+16f(x) = -\dfrac{1}{\sqrt{x^2+16}}.

Q17Stage 2Past exam · 2016A

Find the function y=y(x)y=y(x) that satisfies y(1)=4y(1)=4 and

dydx=15x2+4x+3y\begin{equation*} \diff{y}{x} = \frac{15x^2 + 4x + 3}{y} \end{equation*}
Hint

Be careful about signs.

Answer

y=10x3+4x2+6x−4y = \sqrt{10x^3 + 4x^2 + 6x - 4}

Full solution

This is a separable differential equation that we solve in the usual way. Cross-multiplying and integrating,

y dy=(15x2+4x+3) dx∫y dy=∫(15x2+4x+3) dxy22=5x3+2x2+3x+C.\begin{align*}y\,\dee{y} &= (15x^2 + 4x + 3) \,\dee{x} \\ \int y\,\dee{y} &= \int (15x^2 + 4x + 3) \,\dee{x} \\ \frac{y^2}{2} &= 5x^3 + 2x^2 + 3x + C.\end{align*}

Plugging in x=1x=1 and y=4y=4 gives 422=5+2+3+C,\frac{4^2}2 = 5+2+3+C, and so C=−2C=-2. Therefore

y22=5x3+2x2+3x−2y2=10x3+4x2+6x−4\begin{align*}\frac{y^2}{2} &= 5x^3 + 2x^2 + 3x - 2 \\ y^2 &= 10x^3 + 4x^2 + 6x - 4\end{align*}

This leaves us with two possible functions for yy:

y=10x3+4x2+6x−4ORy=−10x3+4x2+6x−4\begin{align*}y = \sqrt{10x^3 + 4x^2 + 6x - 4} &\qquad\text{OR}\qquad y = - \sqrt{10x^3 + 4x^2 + 6x - 4}\end{align*}

When x=1x=1, y=4y=4. This only fits the first equation, so

y=10x3+4x2+6x−4\begin{align*}y &= \sqrt{10x^3 + 4x^2 + 6x - 4}\end{align*}
Q18Stage 2Past exam · 2002A

Find the solution y(x)y(x) of y′=x3yy'=x^3y with y(0)=1y(0)=1.

Hint

Be careful about signs. If log⁡∣y∣=F\log|y|=F, then ∣y∣=eF|y|=e^F. Since you should give your answer as an explicit function y(x)y(x), you need to decide whether y=eFy=e^F or y=−eFy=-e^F.

Answer

y(x)=ex4/4y(x) = e^{x^4/4}

Full solution

The given differential equation is separable and we solve it accordingly.

dydx=x3ydyy=x3 dx∫dyy=∫x3 dxlog⁡∣y∣=x44+C∣y∣=ex4/4+C=ex4/4eC\begin{align*}\diff{y}{x}&=x^3y \\\frac{\dee{y}}{y}&=x^3\,\dee{x}\\ \int\frac{\dee{y}}{y}&=\int x^3\,\dee{x}\\ \log|y|&=\frac{x^4}{4}+C\\ |y|&=e^{x^4/4+C}=e^{x^4/4}e^C\end{align*}

We are told that y=1y=1 when x=0x=0. That is, 1=e0eC1=e^0e^C, so eC=1e^C=1. That is, C=0C=0.

∣y∣=ex4/4\begin{align*}|y|&=e^{x^4/4}\end{align*}

This leaves us with two potential functions:

y=ex4/4ORy=−ex4/4\begin{align*}y=e^{x^4/4}&\qquad\text{OR}\qquad y=-e^{x^4/4}\end{align*}

The first is always positive, and the second is always negative. Since y=1y=1 (a positive number) when x=0x=0, we see

y=ex4/4\begin{align*}y&=e^{x^4/4}\end{align*}
Q19Stage 2Past exam · 2014A

Solve the initial value problem

xdydx+y=y2y(1)=−1\begin{equation*} x\diff{y}{x} + y = y^2\qquad y(1) = -1 \end{equation*}
Hint

Move the yy from the left hand side to the right hand side, then use partial fractions to integrate.

Be careful about the signs. Remember that we need y=−1y=-1 when x=1x=1. This suggests how to deal with absolute values.

Answer

y=11−2xy=\frac{1}{1-2x}

Full solution

This is a separable differential equation, even if it doesn't quite look like it. First move the yy from the left hand side to the right hand side.

xdydx+y=y2xdydx=y2−y=y(y−1)dyy(y−1)=dxx\begin{align*}x\diff{y}{x}+y&=y^2\\ x\diff{y}{x} &=y^2-y =y(y-1) \\ \frac{\dee{y}}{y(y-1)} &=\frac{\dee{x}}{x}\end{align*}

Using the method of partial fractions, we see 1y(y−1)=1y−1−1y.\frac{1}{y(y-1)} = \frac{1}{y-1}-\frac{1}{y}.

(1y−1−1y) dy=dxx∫(1y−1−1y) dy=∫dxxlog⁡∣y−1∣−log⁡∣y∣=log⁡∣x∣+Clog⁡∣y−1∣∣y∣=log⁡∣x∣+C\begin{align*}\left(\frac{1}{y-1}-\frac{1}{y}\right)\,\dee{y} &=\frac{\dee{x}}{x}\\ \int\left(\frac{1}{y-1}-\frac{1}{y}\right)\,\dee{y} &=\int\frac{\dee{x}}{x}\\ \log|y-1|-\log|y| &=\log |x| + C \\ \log\frac{|y-1|}{|y|}&=\log|x|+C \tag{$*$}\end{align*}

To determine CC we set x=1x=1 and y=−1y=-1.

log⁡∣−2∣∣−1∣=log⁡∣1∣+Clog⁡2=C\begin{align*}\log\frac{|-2|}{|-1|}&=\log|1|+C\\ \log 2&=C\end{align*}

Returning to (∗*),

log⁡∣y−1∣∣y∣=log⁡∣x∣+log⁡2log⁡∣y−1y∣=log⁡∣2x∣∣y−1y∣=∣2x∣\begin{align*}\log\frac{|y-1|}{|y|}&=\log|x|+\log 2\\ \log\left|\frac{y-1}{y}\right|&=\log|2x|\\ \left|\frac{y-1}{y}\right|&=|2x|\end{align*}

As y(1)=−1y(1)=-1 is an initial condition, we have that x≥1x\ge 1 and ∣2x∣=2x|2x|=2x. For x=1x=1, we have y=−1y=-1. So at least for xx near 11, we have yy near −1-1, so that y−1y\frac{y-1}{y} is positive and we may drop the absolute value signs. There remains the possibility that y(x)−1y(x)\frac{y(x)-1}{y(x)} changes sign for some larger x>1x>1. For now, we will simply ignore that possibility. At the end, we will explicitly check that the y(x)y(x) we come up with really does satisfy the differential equation xdydx+y=y2x\diff{y}{x}+y=y^2 and the initial condition y(1)=−1y(1)=-1.

y−1y=2xy−1=2xyy−2xy=1y(1−2x)=1y=11−2x\begin{align*}\frac{y-1}{y}&=2x\\ y-1&=2xy\\ y-2xy&=1\\ y(1-2x)&=1\\ y&=\frac{1}{1-2x}\end{align*}

As a check, we compute:

xdydx+y=xddx{11−2x}+y=x2(1−2x)2+11−2x=2x+(1−2x)(1−2x)2=1(1−2x)2=y2\begin{align*}x\diff{y}{x}+y&=x\diff{}{x}\left\{\frac{1}{1-2x}\right\}+y\\ &=x\frac{2}{(1-2x)^2}+\frac{1}{1-2x}\\ &=\frac{2x+(1-2x)}{(1-2x)^2}\\ &=\frac{1}{(1-2x)^2}\\ &=y^2\end{align*}

So, our differential equation is satisfied. Furthermore:

y(1)=11−2×1=−1\begin{align*}y(1)&=\frac{1}{1-2\times 1}=-1\end{align*}

as desired. This confirms that our solution is correct.

Q20Stage 2Past exam · 2012A

A function f(x)f(x) is always positive, has f(0)=ef(0)=e and satisfies f′(x)=x f(x)f'(x) = x\,f(x) for all xx. Find this function.

Hint

The unknown function f(x)f(x) satisfies an equation that involves the derivative of ff.

Answer

f(x)=e⋅ex2/2f(x) = e\cdot e^{x^2/2}

Full solution

The unknown function f(x)f(x) satisfies an equation that involves the derivative of ff. That means we're in differential equation territory. Specifically, we are told that y=f(x)y=f(x) obeys the separable differential equation dydx=xy\diff{y}{x}=xy.

dydx=xydyy=x dx∫dyy=∫x dxlog⁡∣y∣=x22+C\begin{align*}\diff{y}{x} &= xy\\ \frac{\dee{y}}{y} &= x\,\dee{x} \\ \int \frac{\dee{y}}{y} &= \int x\, \dee{x} \\ \log|y| &=\frac{x^2}{2} + C\end{align*}

To determine CC we set x=0x=0 and y=ey=e.

log⁡e=022+C1=C\begin{align*}\log e &=\frac{0^2}{2}+ C\\ 1&= C\end{align*}

So, the solution is

log⁡∣y∣=x22+1\begin{align*}\log|y|&= \frac{x^2}{2} + 1\end{align*}

We are told that y=f(x)>0y=f(x)>0, so may drop the absolute value signs.

log⁡y=x22+1y=e1+12x2=e⋅ex2/2\begin{align*}\log y &= \frac{x^2}{2}+1 \\ y&=e^{1+\frac{1}{2}x^2}=e\cdot e^{x^2/2}\end{align*}
Q21Stage 2Past exam · M105 2013A

Solve the following initial value problem:

dydx=1(x2+x)yy(1)=2\begin{align*} \diff{y}{x} =\frac{1}{(x^2+x)y}\qquad y(1)=2 \end{align*}
Hint

Try guessing the partial fractions expansion of 1x(x+1)\dfrac{1}{x(x+1)}.

Since x=1x=1 is in the domain and x=0x=0 is not, you may assume x>0x>0 for all xx in the domain.

Answer

y(x)=4+2log⁡2xx+1y(x)=\sqrt{4+2\log\frac{2x}{x+1}}. Note that, to satisfy y(1)=2y(1)=2, we need the positive square root.

Full solution

This is a separable differential equation.

dydx=1(x2+x)yy dy=dxx(x+1)\begin{align*}\diff{y}{x} &=\frac{1}{(x^2+x)y}\\ y\,\dee{y} &=\frac{\dee{x}}{x(x+1)}\end{align*}

Using partial fractions decomposition, we find 1x(x+1)=1x−1x+1\frac{1}{x(x+1)} = \frac{1}{x}-\frac{1}{x+1}.

y dy=(1x−1x+1) dx∫y dy=∫(1x−1x+1) dxy22=log⁡∣x∣−log⁡∣x+1∣+C=log⁡∣xx+1∣+C\begin{align*}y\,\dee{y} &=\left(\frac{1}{x}-\frac{1}{x+1}\right)\,\dee{x}\\ \int y\,\dee{y} &=\int\left(\frac{1}{x}-\frac{1}{x+1}\right)\,\dee{x}\\ \frac{y^2}{2}&=\log|x|-\log|x+1|+C = \log\left| \frac{x}{x+1}\right|+C\end{align*}

To satisfy the initial condition y(1)=2y(1)=2 we must choose CC to obey

222=log⁡∣11+1∣+C2=log⁡12+CC=2−log⁡12\begin{align*}\frac{2^2}{2} &= \log\left|\frac{1}{1+1}\right|+C\\ 2&=\log\frac12+C\\ C&=2-\log\frac{1}{2}\end{align*}

So,

y22=log⁡∣xx+1∣+2−log⁡12y2=2log⁡∣xx+1∣+4−2log⁡12\begin{align*}\frac{y^2}{2}&= \log\left| \frac{x}{x+1}\right|+2-\log\frac{1}{2}\\ y^2&= 2\log\left| \frac{x}{x+1}\right|+4-2\log\frac{1}{2}\end{align*}

Note that the question specifies that y(1)=2y(1)=2 is an initial condition. So we always have x≥1x\ge 1. Then xx+1\frac{x}{x+1} is positive, and we can drop the absolute values.

y2=2log⁡xx+1+4−2log⁡12\begin{align*}y^2&= 2\log \frac{x}{x+1}+4-2\log\frac{1}{2}\end{align*}

This leaves two options for y(x)y(x): the positive or negative square root of the right hand side above. Since y(1)=2y(1)=2, which is positive, we must choose the positive square root.

y(x)=2(log⁡xx+1−log⁡12+2)=4+2log⁡2xx+1\begin{align*}y(x)&=\sqrt{2\Big(\log\frac{x}{x+1}-\log\frac{1}{2}+2\Big)}\\ &=\sqrt{4+2\log\frac{2x}{x+1}}\end{align*}

You might worry that y(x)y(x) could pass through zero, changing sign, at some x>1x>1. But the differential equation says that dydx=1(x2+x)y\diff{y}{x}=\frac{1}{(x^2+x)y} is positive whenever y>0y>0 and x≥1x\ge 1. So y(x)y(x) is an increasing function whenever y>0y>0 and x≥1x\ge 1. As y(1)=2y(1)=2, we have y(x)≥2y(x)\ge 2 for all x≥1x\ge 1.

Q22Stage 2Past exam · 2015A

Find the solution of the differential equation 1+y2−4tan⁡xy′=sec⁡xy\displaystyle \frac{1+\sqrt{y^2-4}}{\tan x} y' = \frac{\sec x}y that satisfies y(0)=2y(0)=2. You don't have to solve for yy in terms of xx.

Hint

ddx{sec⁡x}=sec⁡xtan⁡x\displaystyle\diff{}{x}\{\sec x\} = \sec x \tan x

Answer

y2+23(y2−4)3/2=2sec⁡x+2\displaystyle y^2+\frac{2}{3}(y^2-4)^{3/2}=2\sec x +2

Full solution

This is a separable differential equation.

1+y2−4tan⁡xdydx=sec⁡xyy[1+y2−4] dy=sec⁡x tan⁡x dx∫y[1+y2−4] dy=∫sec⁡x tan⁡x dx\begin{align*}\frac{1+\sqrt{y^2-4}}{\tan x}\diff{y}{x} &= \frac{\sec x}y\\ y\big[1+\sqrt{y^2-4}\big]\ \dee{y} &= \sec x\ \tan x\ \dee{x} \\ \int y\big[1+\sqrt{y^2-4}\big]\ \dee{y} &= \int \sec x\ \tan x\ \dee{x}\end{align*}

For the integral on the left, we use the substitution u=y2−4u={y^2-4}, 12du=y dy\frac12\dee{u}=y\,\dee{y}.

12∫(1+u) du=sec⁡x+C12(u+23u3/2)=sec⁡x+C12(y2−4+23(y2−4)3/2)=sec⁡x+Cy2+23(y2−4)3/2=2sec⁡x+2C+4\begin{align*}\frac{1}{2}\int \big(1+\sqrt{u}\big)\,\dee{u}&=\sec x +C \\\frac{1}{2} \left(u+\frac{2}{3}u^{3/2}\right)&=\sec x +C \\\frac{1}{2} \left(y^2-4+\frac{2}{3}(y^2-4)^{3/2}\right)&=\sec x +C \\ y^2+\frac{2}{3}(y^2-4)^{3/2}&=2\sec x +2C+4\end{align*}

To find CC we set x=0x=0 and y=2y=2.

4+234−43=2sec⁡(0)+2C+44=2+2C+42=2C+4\begin{align*}4+\frac{2}{3}\sqrt{4-4}^3&=2\sec (0)+2C+4\\ 4&=2+2C+4\\ 2&=2C+4\end{align*}

So,

y2+23(y2−4)3/2=2sec⁡x+2\begin{align*}y^2+\frac{2}{3}(y^2-4)^{3/2}&=2\sec x +2\end{align*}
Q23Stage 2Past exam · 1996A

The fish population in a lake is attacked by a disease at time t=0t=0, with the result that the size P(t)P(t) of the population at time t≥0t\ge 0 satisfies

dPdt=−kP\begin{align*} \diff{P}{t}=-k\sqrt{P} \end{align*}

where kk is a positive constant. If there were initially 90,000 fish in the lake and 40,000 were left after 6 weeks, when will the fish population be reduced to 10,000?

Hint

The general solution to the differential equation will contain the constant kk and one other constant. They are determined by the data given in the question.

Answer

12 weeks12\text{ weeks}

Full solution

The given differential equation is separable and we solve it accordingly.

dPdt=−kPdPP=−k dt∫dPP=∫−k dt2P=−kt+C\begin{align*}\diff{P}{t}&=-k\sqrt{P}\\ \frac{\dee{P}}{\sqrt{P}}&=-k\,\dee{t}\\ \int\frac{\dee{P}}{\sqrt{P}}&=\int-k\,\dee{t}\\ 2\sqrt{P}&=-kt+C\end{align*}

At t=0t=0, P=90,000P=90,000 so

290,000=−k×0+CC=2×300=600\begin{align*}2\sqrt{90,000}&=-k\times 0+C\\ C&=2\times 300=600\end{align*}

Therefore,

2P=−kt+600\begin{align*}2\sqrt{P}&=-kt+600\tag{$*$}\end{align*}

Now, we find kk. Let tt be measured in weeks. Then when t=6t=6, P=40,000P=40,000.

240,000=−6k+6002⋅200=−6k+600k=2006=1003\begin{align*}2\sqrt{40,000}&=-6k+600\\ 2\cdot 200&=-6k+600\\ k&=\frac{200}{6} = \frac{100}{3}\end{align*}

Substituting our value of kk into (∗*):

2P=−1003t+600\begin{align*}2\sqrt{P}&=-\frac{100}{3}t+600\end{align*}

To find when the population will be 10,000, we set P=10,000P=10,000 and solve for tt.

210,000=−1003t+6002⋅100=−1003t+6001003t=400t=12\begin{align*}2\sqrt{10,000}&=-\frac{100}{3}t+600\\ 2\cdot 100&=-\frac{100}{3}t+600\\ \frac{100}{3}t&=400\\ t&=12\end{align*}

Since we measured tt in weeks when we found kk, we see that in 12 weeks the population will decrease to 10,000 individuals.

Q24Stage 2Past exam · 1996A

An object of mass mm is projected straight upward at time t=0t=0 with initial speed v0v_0. While it is going up, the only forces acting on it are gravity (assumed constant) and a drag force proportional to the square of the object's speed v(t)v(t). It follows that the differential equation of motion is

mdvdt=−(mg+kv2)\begin{align*} m\diff{v}{t}=-(mg+kv^2) \end{align*}

where gg and kk are positive constants. At what time does the object reach its highest point?

Hint

  • When you're solving the differential equation, you should have an integral that you can massage to look something like arctangent.

  • What is the velocity of the object at its highest point?

  • Your final answer will depend on the (unspecified) constants v0v_0, mm, gg and kk.

Answer

t=mkgarctan⁡(kmg v0)t=\displaystyle\sqrt{\frac{m}{kg}}\arctan \left(\sqrt{\frac{k}{mg}}\,v_0\right)

Full solution

The given differential equation is separable and we solve it accordingly.

mdvdt=−(mg+kv2)mmg+kv2 dv=−dt∫mmg+kv2 dv=∫−dt\begin{align*}m\diff{v}{t}&=-(mg+kv^2)\\ \frac{m}{mg+kv^2}\,\dee{v}&=-\dee{t}\\ \int\frac{m}{mg+kv^2}\,\dee{v}&=\int-\dee{t}\end{align*}

The left integral looks something like the antiderivative of arctangent. Let's factor out that mgmg from the denominator.

1mg∫m1+kmgv2 dv=−t+C1g∫11+(kmgv)2 dv=−t+C\begin{align*}\frac{1}{mg}\int\frac{m}{1+\frac{k}{mg}v^2}\,\dee{v}&=-t+C\\ \frac{1}{g}\int\frac{1}{1+\left(\sqrt{\frac{k}{mg}}v\right)^2}\,\dee{v}&=-t+C\end{align*}

Now it looks even more like the derivative of arctangent. We can guess the antiderivative from here, or use the substitution u=kmgvu=\sqrt{\frac{k}{mg}}v, du=kmg dv\dee{u}=\sqrt{\frac{k}{mg}}\,\dee{v}.

1gmgkarctan⁡(kmgv)=−t+Cmgkarctan⁡(kmgv)=−t+C\begin{align*}\frac{1}{g}\sqrt{\frac{mg}{k}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=-t+C\\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=-t+C\tag{$*$}\end{align*}

At t=0t=0, v=v0v=v_0, so:

mgkarctan⁡(kmgv0)=C\begin{align*}\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)&=C\end{align*}

Plug CC into (∗)(*).

mgkarctan⁡(kmgv)=mgkarctan⁡(kmgv0)−t\begin{align*}\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t\end{align*}

At its highest point, the object has velocity v=0v=0. This happens when tt obeys:

mgkarctan⁡(kmg0)=mgkarctan⁡(kmgv0)−t0=mgkarctan⁡(kmgv0)−tt=mgkarctan⁡(kmgv0)\begin{align*}\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}0\right)&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t\\ 0&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t\\ t&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)\end{align*}
Q25Stage 2Past exam · 1996D

A motor boat is traveling with a velocity of 40 ft/sec when its motor shuts off at time t=0t=0. Thereafter, its deceleration due to water resistance is given by

dvdt=−k v2\begin{align*} \diff{v}{t}=-k\,v^2 \end{align*}

where kk is a positive constant. After 10 seconds, the boat's velocity is 20 ft/sec.

  1. What is the value of kk?

  2. When will the boat's velocity be 5 ft/sec?

Hint

The general solution to the differential equation will contain the constant kk and one other constant. They are determined by the data given in the question.

Answer

(a) k=1400k=\frac{1}{400} (b) t=70sect=70\mathrm{sec}

Full solution

(a) The given differential equation is separable and we solve it accordingly.

dvdt=−k v2−dvv2=k dt∫−dvv2=∫k dt1v=kt+C\begin{align*}\diff{v}{t}&=-k\,v^2\\ -\frac{\dee{v}}{v^2}&=k\,\dee{t}\\ \int- \frac{\dee{v}}{v^2}&=\int k\,\dee{t}\\ \frac{1}{v}&=kt+C\end{align*}

At t=0t=0, v=40v=40 so

140=k×0+CC=140\begin{align*}\frac{1}{40}&=k\times 0+ C\\ C&=\frac{1}{40}\end{align*}

Therefore,

v(t)=1kt+C=1kt+1/40=4040kt+1\begin{align*}v(t)&=\frac{1}{kt+C}=\frac{1}{kt+1/40}=\frac{40}{40kt+1}\tag{$*$}\end{align*}

The constant of proportionality kk is determined by

v(10)=2020=4040k×10+112=1400k+1400k+1=2k=1400\begin{align*}v(10)&=20\\ 20&=\frac{40}{40k\times 10+1}\\ \frac{1}{2}&=\frac{1}{400k+1}\\ 400k+1&=2 \\ k&=\frac{1}{400}\end{align*}

(b) Subbing in the value of kk to (∗)(*),

v(t)=4040kt+1=40t/10+1\begin{align*}v(t)&=\frac{40}{40kt+1}=\frac{40}{t/10+1}\end{align*}

We want to know the value of tt that gives v(t)=5v(t)=5.

5=40t/10+1t10+1=8t=70 sec\begin{align*}5&=\frac{40}{t/10+1}\\ \frac{t}{10}+1&=8 \\ t&=70\text{ sec}\end{align*}
Q26Stage 2Past exam · M121 2000A

Consider the initial value problem dxdt=k(3−x)(2−x)\diff{x}{t}= k(3-x)(2-x), x(0)=1x(0)=1, where kk is a positive constant. (This kind of problem occurs in the analysis of certain chemical reactions.)

  1. Solve the initial value problem. That is, find xx as a function of tt.

  2. What value will x(t)x(t) approach as tt approaches +∞+\infty.

Hint

The method of partial fractions will help you integrate.

To solve x−ax−b=Y\frac{x-a}{x-b}=Y for xx, move the terms containing xx out of the denominator, then gather them on one side of the equals sign and factor out the xx.

x−ax−b=Yx−a=Y(x−b)=Yx−Ybx−Yx=a−Ybx(1−Y)=a−Ybx=a−Yb1−Y\begin{align*} \frac{\textcolor{red}{x}-a}{\textcolor{red}{x}-b}&=Y\\ \textcolor{red}{x}-a&=Y(\textcolor{red}{x}-b)=Y\textcolor{red}{x}-Yb\\ \textcolor{red}{x}-Y\textcolor{red}{x}&=a-Yb\\ \textcolor{red}{x}(1-Y)&=a-Yb\\ \textcolor{red}{x}=\frac{a-Yb}{1-Y} \end{align*}

To find the limit, you can avoid l'H^opital's rule using some clever algebra–but you can also just use l'H^opital's rule.

Answer

(a) x(t)=3−4ekt1−2ektx(t)=\dfrac{3-4e^{kt}}{1-2e^{kt}} (b) As t→∞t\rightarrow\infty, x→2x\rightarrow 2.

Full solution

(a) The given differential equation is separable and we solve it accordingly.

dxdt=k(3−x)(2−x)dx(x−2)(x−3)=kdt\begin{align*}\diff{x}{t}&= k(3-x)(2-x)\\ \frac{\dee{x}}{(x-2)(x-3)}&= k\dee{t}\end{align*}

Using the method of partial fractions, we find 1(x−2)(x−3)=1x−3−1x−2\frac{1}{(x-2)(x-3)}=\frac{1}{x-3}-\frac{1}{x-2}.

∫[1x−3−1x−2] dx=∫kdtlog⁡∣x−3∣−log⁡∣x−2∣=kt+Clog⁡∣x−3x−2∣=kt+C∣x−3x−2∣=ekt+C=ekteCx−3x−2=Dekt\begin{align*}\int\Big[\frac{1}{x-3}-\frac{1}{x-2}\Big]\ \dee{x}&= \int k\dee{t}\\ \log|x-3|-\log|x-2|&=kt+C\\ \log\left|\frac{x-3}{x-2}\right|&=kt+C\\ \left|\frac{x-3}{x-2}\right|&=e^{kt+C}=e^{kt}e^C\\ \frac{x-3}{x-2}&=De^{kt}\end{align*}

where D=±eCD=\pm e^C. When t=0t=0, x=1x=1, forcing

1−31−2=De0D=2\begin{align*}\frac{1-3}{1-2}&=De^{0}\\ D&=2\end{align*}

Hence

x−3x−2=2ektx−3=2ekt(x−2)x−2ektx=3−4ektx(t)=3−4ekt1−2ekt\begin{align*}\frac{x-3}{x-2}&=2e^{kt}\\ x-3&=2e^{kt}(x-2)\\ x-2e^{kt}x&=3-4e^{kt}\\ x(t)&=\frac{3-4e^{kt}}{1-2e^{kt}}\end{align*}

(b) To evaluate the limit, we could use l'H^opital's rule, but we could also just multiply the numerator and denominator by e−kte^{-kt}. Note lim⁡t→∞e−tk=0\lim\limits_{t \to \infty} e^{-tk}=0.

lim⁡t→∞x(t)=lim⁡t→∞3−4ekt1−2ekt⏟num→−∞den→−∞=lim⁡t→∞3−4ekt1−2ekt⋅e−kte−kt=lim⁡t→∞3e−kt−4e−kt−2=0−40−2=2\begin{align*} \lim_{t\rightarrow\infty}x(t) &=\lim_{t\rightarrow\infty}\underbrace{\frac{3-4e^{kt}}{1-2e^{kt}}}_{\atp{\mathrm{num}\to-\infty}{\mathrm{den}\to-\infty}} =\lim_{t \to \infty} \frac{3-4e^{kt}}{1-2e^{kt}}\cdot\frac{e^{-kt}}{e^{-kt}} =\lim_{t\rightarrow\infty}\frac{3e^{-kt}-4}{e^{-kt}-2} =\frac{0-4}{0-2} =2 \end{align*}
Q27Stage 2Past exam · 1997D

The quantity P=P(t)P=P(t), which is a function of time tt, satisfies the differential equation

dPdt=4P−P2\begin{align*} \diff{P}{t}=4P-P^2 \end{align*}

and the initial condition P(0)=2P(0)=2.

  1. Solve this equation for P(t)P(t).

  2. What is PP when t=0.5t=0.5? What is the limiting value of PP as tt becomes large?

Hint

Be careful about signs.

Part (a) has some algebraic similarities to Question 26.

Answer

(a) P=41+e−4tP=\dfrac{4}{1+e^{-4t}} (b) At t=12t=\dfrac{1}{2}, P≈3.523P\approx 3.523. As t→∞t\rightarrow\infty, P→4P\rightarrow 4.

Full solution

(a) The given differential equation is separable and we solve it accordingly.

dPdt=4P−P2dP4P−P2=dtdPP(4−P)=dt\begin{align*}\diff{P}{t}&=4P-P^2\\ \frac{\dee{P}}{4P-P^2}&=\dee{t}\\ \frac{\dee{P}}{P(4-P)}&=\dee{t}\end{align*}

Using the method of partial fractions, we see 1P(4−P)=1/4P+1/44−P\frac{1}{P(4-P)}=\frac{1/4}{P} + \frac{1/4}{4-P}.

14[1P+14−P]dP=dt∫14[1P+14−P]dP=∫dt14[log⁡∣P∣−log⁡∣4−P∣]=t+C\begin{align*}\frac{1}{4}\Big[\frac{1}{P}+\frac{1}{4-P}\Big]\dee{P}&=\dee{t}\\ \int \frac{1}{4}\Big[\frac{1}{P}+\frac{1}{4-P}\Big]\dee{P}&=\int\dee{t}\\ \frac{1}{4}\big[\log|P|-\log|4-P|\Big]&=t+C\end{align*}

When t=0t=0, P=2P=2, so 14[log⁡∣2∣−log⁡∣2∣]=C  ⟹  C=0\frac{1}{4}\big[\log|2|-\log|2|\big]=C\implies C=0. So,

14log⁡∣P4−P∣=t\begin{align*}\frac{1}{4}\log\Big|\frac{P}{4-P}\Big|&=t\end{align*}

At time t=0t=0, P4−P=1>0\frac{P}{4-P}=1>0. The ratio may not change sign at any finite time, because this could only happen if at some finite time PP took either the value 0 or the value 4. But at this time t=14log⁡∣P4−P∣t=\frac{1}{4}\log\big|\frac{P}{4-P}\big| would have to be infinite. So P4−P>0\frac{P}{4-P}>0 for all time and:

14log⁡P4−P=tlog⁡P4−P=4tP4−P=e4tP=(4−P)e4tP+Pe4t=4e4tP=4e4t1+e4t=41+e−4t\begin{align*}\frac{1}{4}\log \frac{P}{4-P} &=t\\ \log \frac{P}{4-P} &=4t\\ \frac{P}{4-P} &=e^{4t} \\ P&=(4-P)e^{4t}\\ P+Pe^{4t}&=4e^{4t} \\ P&=\frac{4e^{4t}}{1+e^{4t}}=\frac{4}{1+e^{-4t}}\end{align*}

(b) At t=12t=\frac{1}{2}, P=41+e−2≈3.523P=\frac{4}{1+e^{-2}}\approx 3.523.

lim⁡t→∞P(t)=lim⁡t→∞41+e−4t=41+0=4\lim_{t \to \infty} P(t)=\lim_{t \to \infty}\frac{4}{1+e^{-4t}} = \frac{4}{1+0}=4
Q28Stage 2Past exam · 1998A

An object moving in a fluid has an initial velocity vv of 400 m/min. The velocity is decreasing at a rate proportional to the square of the velocity. After 1 minute the velocity is 200 m/min.

  1. Give a differential equation for the velocity v=v(t)v=v(t) where tt is time.

  2. Solve this differential equation.

  3. When will the object be moving at 50 m/min?

Hint

The general solution to the differential equation will contain a constant of proportionality and one other constant. They are determined by the data given in the question.

Answer

(a) dvdt=−kv2\displaystyle \diff{v}{t}=-kv^2 (b) v=400t+1\displaystyle v=\frac{400}{t+1} (c) t=7t=7

Full solution

  1. The rate of change of speed at time tt is −kv(t)2-kv(t)^2 for some constant of proportionality kk (to be determined–but we assume it is positive, since the speed is decreasing). So v(t)v(t) obeys the differential equation dvdt=−kv2\diff{v}{t}=-kv^2 .

  2. The equation dvdt=−kv2\diff{v}{t}=-kv^2 is a separable differential equation, which we can solve in the usual way.

    dvdt=−kv2dv−v2=kdt∫−dvv2=∫kdt1v=kt+C\begin{align*}\diff{v}{t}&=-kv^2\\ \frac{\dee{v}}{-v^2}&=k\dee{t}\\ \int -\frac{\dee{v}}{v^2} &=\int k\dee{t}\\ \frac{1}{v}&=kt+C\end{align*}

    At time t=0t=0, v=400v=400, so C=1400C=\frac{1}{400}. Then:

    1v=kt+1400\begin{align*}\frac{1}{v}&=kt+\frac{1}{400}\tag{$*$}\end{align*}

    At time t=1t=1, v=200v=200, so

    1200=k+1400k=1400\begin{align*}\frac{1}{200}&=k+\frac{1}{400}\\ k&=\frac{1}{400}\end{align*}

    Therefore, from (∗*),

    1v=t400+1400=t+1400v=400t+1\begin{align*}\frac{1}{v}&=\frac{t}{400}+\frac{1}{400} = \frac{t+1}{400}\\ v&=\frac{400}{t+1}\end{align*}

  3. To find when the speed is 50, we set v=50v=50 in the equation from (b) and solve for tt.

    50=400t+150(t+1)=400t+1=8t=7\begin{align*} 50&=\frac{400}{t+1}\\ 50(t+1)&=400\\ t+1&=8\\ t&=7 \end{align*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q29Stage 3Past exam · 1997A

An investor places some money in a mutual fund where the interest is compounded continuously and where the interest rate fluctuates between 4%4\% and 8%8\%. Assume that the amount of money B=B(t)B=B(t) in the account in dollars after tt years satisfies the differential equation

dBdt=(0.06+0.02sin⁡t)B\begin{align*} \diff{B}{t}=\big(0.06+0.02\sin t\big)B \end{align*}

  1. Solve this differential equation for BB as a function of tt.

  2. If the initial investment is $1000\char36 1000, what will the balance be at the end of two years?

Hint

You do not need to know anything about investing or continuous compounding to do this problem. You are given the differential equation explicitly. The whole first sentence is just window dressing.

Answer

(a) B(t)=C e0.06t−0.02cos⁡tB(t)=C\,e^{0.06 t-0.02\cos t} with the arbitrary constant C≥0C\ge 0. (b) $1159.89\char36 1159.89

Full solution

(a) The given differential equation is separable and we solve it accordingly.

dBdt=(0.06+0.02sin⁡t)BdBB=(0.06+0.02sin⁡t) dt∫dBB=∫(0.06+0.02sin⁡t) dtlog⁡∣B(t)∣=0.06t−0.02cos⁡t+C′\begin{align*}\diff{B}{t}&=(0.06+0.02\sin t)B\\ \frac{\dee{B}}{B}&=(0.06+0.02\sin t)\,\dee{t} \\ \int \frac{\dee{B}}{B}&=\int (0.06+0.02\sin t)\,\dee{t} \\ \log |B(t)|&= 0.06 t-0.02\cos t +C'\end{align*}

Since B(t)B(t) is our bank account balance and we're not withdrawing money, B(t)B(t) is positive, so we can drop the absolute value signs.

log⁡B(t)=0.06t−0.02cos⁡t+C′B(t)=e0.06t−0.02cos⁡teC′B(t)=Ce0.06t−0.02cos⁡t\begin{align*}\log B(t)&= 0.06 t-0.02\cos t +C'\\ B(t)&= e^{0.06 t-0.02\cos t}e^{C'}\\ B(t)&=Ce^{0.06 t-0.02\cos t}\end{align*}

for arbitrary constants C′C' and C=eC′≥0C=e^{C'}\ge 0.

Remark: the function B(t)=0B(t)=0 obeys the differential equation so that C=0C=0 is allowed, even though it is not of the form C=eC′C=e^{C'}. This seeming discrepancy arose because, in our very first step of part (a), we divided both sides of the differential equation by BB, which is only allowable if B≠0B\neq 0. So, in this step, we implicitly assumed BB was nonzero.

(b) We are told that B(0)=1000B(0)=1000. This allows us to find CC.

1000=B(0)=Ce0−0.02cos⁡0=Ce−0.02C=1000e0.02\begin{align*}1000=B(0)&=Ce^{0-0.02\cos 0}=Ce^{-0.02}\\ C&=1000e^{0.02}\end{align*}

So, when t=2t=2,

B(2)=p1000e0.02⏟Ce0.06×2−0.02cos⁡2=$1159.89\begin{align*}B(2)&=\underbrace{\vphantom{p}1000e^{0.02}}_{C} e^{0.06\times 2-0.02\cos 2}=\char36 1159.89\end{align*}

rounded to the nearest cent.

Note that cos⁡2\cos 2 is the cosine of 2 radians, cos⁡2≈−0.416\cos 2 \approx -0.416.

Q30Stage 3Past exam · M105 2014A

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem B′(t)=aB−mB'(t) = aB - m for t≥0t \ge 0, with B(0)=B0B(0) = B_0 . The constant aa reflects the annual interest rate, mm is the annual rate of withdrawal, and B0B_0 is the initial balance in the account.

  1. Solve the initial value problem with a=0.02a = 0.02 and B(0)=B0=$30,000B(0) = B_0 = \char36 30,000. Note that your answer depends on the constant mm.

  2. If a=0.02a = 0.02 and B(0)=B0=$30,000B(0) = B_0 = \char36 30,000, what is the annual withdrawal rate mm that ensures a constant balance in the account?

Hint

Again, you do not need to know anything about investing to do this problem. You are given the differential equation explicitly.

Answer

(a) B(t)={30000−50m}et/50+50mB(t) = \left\{30000-50m\right\} e^{t/50} + 50m (b) $600\char36 600

Full solution

(a) The given differential equation is separable and we could solve it accordingly. In fact we have already done so. If we rewrite the equation in the form

dBdt=a(B−ma)\begin{align*} \diff{B}{t} = a\Big(B-\frac{m}{a}\Big) \end{align*}

it is of the form covered by Theorem 2.4.4 in the CLP-2 text. So that theorem tells us that the solution is

B(t)=(B(0)−ma)eat+ma\begin{align*} B(t) = \left(B(0)-\frac{m}{a}\right) e^{at} + \frac{m}{a} \end{align*}

In this problem we are told that a=0.02=150a=0.02=\frac{1}{50}, so

B(t)={B(0)−50m}et/50+50m={30000−50m}et/50+50m\begin{align*} B(t) = \left\{B(0)-50m\right\} e^{t/50} + 50m = \left\{30000-50m\right\} e^{t/50} + 50m \end{align*}

(b) The solution of part (a) is independent of time if and only if 30000−50m=030000-50m=0. So we need

m=3000050=$600\begin{align*} m= \frac{30000}{50} = \char36 600 \end{align*}
Q31Stage 3Past exam · M121 1999A

A certain continuous function y=y(x)y=y(x) satisfies the integral equation

y(x)=3+∫0x(y(t)2−3y(t)+2)sin⁡t dt\begin{equation} y(x)=3+\int_0^x\big(y(t)^2-3y(t)+2\big)\sin t\ \dee{t} \tag{$*$}\end{equation}

for all xx in some open interval containing 00. Find y(x)y(x) and the largest interval for which (∗)(*) holds.

Hint

Differentiate the given integral equation. Plugging in x=0x=0 gives you y(0)y(0).

Answer

y(x)=4−e1−cos⁡x2−e1−cos⁡xy(x)=\dfrac{4-e^{1-\cos x}}{2-e^{1-\cos x}}. The largest allowed interval is

−arccos⁡(1−log⁡2)<x<arccos⁡(1−log⁡2)\begin{equation*} -\arccos(1-\log 2)<x<\arccos(1-\log 2) \end{equation*}

or, roughly, −1.259<x<1.259- 1.259 < x < 1.259.

Full solution

What we're given is an equation relating yy to the integral of a function of yy. What we know how to solve is an equation relating the derivative of yy to a function of yy. We can create this by differentiating the given integral equation. By the Fundamental Theorem of Calculus, part 1:

y′(x)=ddx{∫0x(y(t)2−3y(t)+2)sin⁡t dt}=(y(x)2−3y(x)+2)sin⁡x\begin{equation*} y'(x)=\diff{}{x}\left\{\int_0^x\big(y(t)^2-3y(t)+2\big)\sin t\ \dee{t}\right\} =\big(y(x)^2-3y(x)+2\big)\sin x \end{equation*}

So y(x)y(x) satisfies the differential equation y′=(y2−3y+2)sin⁡x=(y−2)(y−1)sin⁡xy'=\big(y^2-3y+2\big)\sin x =(y-2)(y-1)\sin x and the initial equation y(0)=3y(0)=3 (just substitute x=0x=0 into (∗)(*)). For y≠1,2y\ne 1,2:

dydx=(y−2)(y−1)sin⁡xdy(y−2)(y−1)=sin⁡x dx∫dy(y−2)(y−1)=∫sin⁡x dx\begin{align*}\diff{y}{x}&=(y-2)(y-1)\sin x\\ \frac{\dee{y}}{(y-2)(y-1)}&=\sin x\ \dee{x} \\\int \frac{\dee{y}}{(y-2)(y-1)}&=\int \sin x\ \dee{x}\end{align*}

Using the method of partial fractions, we see 1(y−2)(y−1)=1y−2−1y−1\frac{1}{(y-2)(y-1)}=\frac{1}{y-2}-\frac{1}{y-1}.

∫[1y−2−1y−1]dy=∫sin⁡x dxlog⁡∣y−2∣−log⁡∣y−1∣=−cos⁡x+clog⁡∣y−2y−1∣=−cos⁡x+c∣y−2y−1∣=ec−cos⁡x\begin{align*}\int\Big[\frac{1}{y-2}-\frac{1}{y-1}\Big]\dee{y}&=\int\sin x\ \dee{x}\\ \log|y-2|-\log|y-1|&=-\cos x+c\\ \log\left|\frac{y-2}{y-1}\right|&=-\cos x+c\\ \left|\frac{y-2}{y-1}\right|&= e^{c-\cos x}\end{align*}

The condition y(0)=3y(0)=3 forces ∣3−23−1∣=ec−1\big|\frac{3-2}{3-1}\big|= e^{c-1} or ec=12ee^c=\frac{1}{2} e, hence

∣y−2y−1∣=12e1−cos⁡x\begin{align*}\left|\frac{y-2}{y-1}\right|&=\frac{1}{2} e^{1-\cos x}\end{align*}

Observe that, when x=0x=0, y−2y−1=12>0\frac{y-2}{y-1}=\frac{1}{2}>0. Furthermore 12e1−cos⁡x\frac{1}{2} e^{1-\cos x}, and hence ∣y−2y−1∣\big|\frac{y-2}{y-1}\big|, can never take the value zero. As y(x)y(x) varies continuously with xx, y(x)y(x) must remain larger than 2. Consquently, y−2y−1\frac{y-2}{y-1} remains positive and we may drop the absolute value signs. Hence

y−2y−1=12e1−cos⁡x\begin{align*}\frac{y-2}{y-1}=\frac{1}{2} e^{1-\cos x}\end{align*}

Solving for yy,

y−2y−1=12e1−cos⁡x2(y−2)=e1−cos⁡x(y−1)2y−4=ye1−cos⁡x−e1−cos⁡xy(2−e1−cos⁡x)=4−e1−cos⁡xy=4−e1−cos⁡x2−e1−cos⁡x\begin{align*}\frac{y-2}{y-1}&=\frac{1}{2} e^{1-\cos x}\\ 2(y-2)&=e^{1-\cos x}(y-1)\\ 2y-4&=ye^{1-\cos x}-e^{1-\cos x}\\ y\big(2-e^{1-\cos x}\big)&=4-e^{1-\cos x} \\ y&=\frac{4-e^{1-\cos x}}{2-e^{1-\cos x}}\end{align*}

To avoid division by zero in the last step, we need

e1−cos⁡x≠21−cos⁡x≠log⁡2cos⁡x≠1−log⁡2\begin{alignat*}{3} e^{1-\cos x}&\neq2\\ 1-\cos x&\neq \log 2\\ \cos x&\neq 1-\log 2 \end{alignat*}

Let L=1−log⁡2L=1-\log 2, for brevity, and note that L>0L>0. (This can be seen by observing 2<e2<e, so, log⁡2<log⁡e=1\log 2<\log e = 1, hence 1−log⁡2>01-\log 2>0.)

Figure from prob_s2.4, line 2

Figure from prob_s2.4, line 2

We know x=0x=0 is in the domain of our function, but the points x=±arccos⁡(L)=±arccos⁡(1−log⁡2)x=\pm \arccos(L) = \pm\arccos(1-\log 2) are not.

Figure from prob_s2.4, line 2

Figure from prob_s2.4, line 2

Therefore, the largest interval for which our answer makes sense is

−arccos⁡(1−log⁡2))>x>arccos⁡(1−log⁡2)-\arccos(1-\log 2))>x>\arccos(1-\log 2)

or approximately −1.259<x<1.259- 1.259 < x < 1.259.

Q32Stage 3Past exam · M121 2001A

A cylindrical water tank, of radius 3 meters and height 6 meters, is full of water when its bottom is punctured. Water drains out through a hole of radius 1 centimeter. If

  • h(t)h(t) is the height of the water in the tank at time tt (in meters) and

  • v(t)v(t) is the velocity of the escaping water at time tt (in meters per second) then

  • Torricelli's law states that v(t)=2gh(t)v(t)=\sqrt{2gh(t)} where g=9.8 m/sec2g=9.8\ {\rm m/sec^2}. Determine how long it takes for the tank to empty.

Hint

Suppose that in a very short time interval dt\dee{t}, the height of water in the tank changes by dh\dee{h} (which is negative). Express in two different ways the volume of water that has escaped during this time interval. Equating the two gives the needed differential equation.

As the water escapes, it forms a cylinder of radius 1 cm.

Answer

180,0003g≈99,591 sec≈27.66 hr180,000 \sqrt{\frac{3}{g}}\approx 99,591\text{ sec} \approx 27.66\text{ hr}

Full solution

Suppose that in a very short time interval dt\dee{t}, the height of water in the tank changes by dh\dee{h} (which is negative). Then in this time interval the amount of the water in the tank decreases by dV=−π(3)2dh\dee{V}=-\pi(3)^2\dee{h}. This must be the same as the amount of water that flows through the hole in this time interval. The water flowing through the hole makes a cylinder of radius 1 cm (that is, 0.01 m) with length v(t)dtv(t)\dee{t}, the distance the water moves out of the hole in dt\dee{t} seconds. So, the amount of water leaving the hole over the time interval dt\dee{t} is π(0.01)2v(t) dt=π(0.01)22gh(t) dt\pi(0.01)^2 v(t)\,\dee{t} =\pi(0.01)^2 \sqrt{2gh(t)}\,\dee{t}.

Figure from prob_s2.4, line 1

Figure from prob_s2.4, line 1

This gives us a separable differential equation. Recall gg is a constant.

−π(3)2dh=π(0.01)22gh(t) dtdhh=−(0.013) ⁣22g dt∫dhh=∫−(0.013) ⁣22g dt2h=−(0.013) ⁣22g t+C\begin{align*}-\pi(3)^2\dee{h}&=\pi(0.01)^2 \sqrt{2gh(t)}\,\dee{t}\\ \frac{\dee{h}}{\sqrt{h}}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,\dee{t}\cr \int\frac{\dee{h}}{\sqrt{h}}&=\int-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,\dee{t}\cr 2\sqrt{h}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,t+C\end{align*}

At time 00, the height is 66, so C=26C=2\sqrt{6} and

2h=−(0.013) ⁣22g t+26\begin{align*}2\sqrt{h}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,t+2\sqrt{6}\end{align*}

We want to know when the height of the water in the tank is 0.

0=−(0.013) ⁣22g t+26(0.013) ⁣22g t=26t=26(0.013) ⁣22g=2(30.01) ⁣23g=180,0003g≈99,591 sec≈27.66 hr\begin{align*}0&=-\Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}\,t+2\sqrt{6}\\ \Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}\,t&=2\sqrt{6}\\ t&=\frac{2\sqrt6}{\Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}}\\ &=2{\Big(\frac{3}{0.01}\Big)}^{\!2} \sqrt{\frac{3}{g}}\\ &=180,000 \sqrt{\frac{3}{g}}\approx 99,591\text{ sec} \approx 27.66\text{ hr}\end{align*}
Q33Stage 3Past exam · 2000D

A spherical tank of radius 6 feet is full of mercury when a circular hole of radius 1 inch is opened in the bottom. How long will it take for all of the mercury to drain from the tank?

Use the value g=32 feet/sec2g=32\ {\rm feet}/{\rm sec}^2. Also use Torricelli's law, which states when the height of mercury in the tank is hh, the speed of the mercury escaping from the tank is v=2ghv=\sqrt{2gh}.

Hint

Sketch the mercury in the tank at time tt, when it has height hh, and also at time t+dtt+\dee{t}, when it has height h+dhh+\dee{h} (with dh<0\dee{h}<0). The difference between those two volumes is the volume of (essentially) a disk of thickness −dh-\dee{h}. Figure out the radius and then the volume of that disk. This volume has to be the same as the volume of mercury that left through the hole in the bottom of the sphere, which runs out in the shape of a cylinder. Toricelli's law tells you what the length of that cylinder is, and from there you can find its volume. Setting the two volumes equal to each other gives the differential equation that determines h(t)h(t).

Answer

t=4×144151252g≈2,394 sec ≈0.665 hr\displaystyle t=\frac{4\times 144}{15} \sqrt{\frac{12^5}{2g}} \approx 2,394\,\text{sec }\approx 0.665\, \text{hr}

Full solution

Suppose that at time tt, the mercury in the tank has height hh, which is between 0 and 12 feet.

Figure from prob_s2.4, line 1930

Figure from prob_s2.4, line 1930

At that time, the top surface of the mercury forms a circular disk of radius 62−(h−6)2\sqrt{6^2-(h-6)^2}. (We found this by applying the Pythagorean Theorem to the triangle in the diagram above. In the diagram, hh is shown as being larger than 6, but the same equation holds for all hh in [0,12][0,12].) Now suppose that in a very short time interval dt\dee{t}, the height of mercury in the tank changes by dh\dee{h} (which is negative). Then in this time interval the amount of the mercury in the tank decreases by −π(62−(h−6)2 )2dh-\pi\big(\sqrt{6^2-(h-6)^2}\ \big)^2\dee{h}. (That's the volume of the red disk in the figure above.) This must be the same as the amount of mercury that flows through the hole in this time interval. The mercury comes out of the hole as a cylinder. Its radius is the radius of the hole, 112\frac{1}{12} foot, and its length is the distance the mercury travels in dt\dee{t} seconds, v(t)dtv(t)\dee{t} feet. So, the volume of escaped mercury is π(112)2v dt=π(112)22gh dt\pi\big(\frac{1}{12}\big)^2 v\,\dee{t} =\pi\big(\frac{1}{12}\big)^2 \sqrt{2gh}\,\dee{t}. This gives us a separable differential equation.

−π(62−(h−6)2 )2dh=π(112)22gh dt−(36−(h2−12h+36) )dh=(112)22gh dt(h2−12h)dh=11442gh dt(h3/2−12h1/2)dh=11442g dt∫(h3/2−12h1/2)dh=∫11442g dth5/25/2−12h3/23/2=11442g t+C\begin{align*}-\pi{\big(\sqrt{6^2-(h-6)^2}\ \big)}^2\dee{h} &=\pi\Big(\frac{1}{12}\Big)^2 \sqrt{2gh}\,\dee{t}\\ -{\big({36-(h^2-12h+36)}\ \big)}\dee{h} &=\Big(\frac{1}{12}\Big)^2 \sqrt{2gh}\,\dee{t}\\ \big(h^2-12h\big)\dee{h} &=\frac{1}{144} \sqrt{2g}\sqrt{h}\,\dee{t}\cr \big(h^{3/2}-12h^{1/2}\big)\dee{h}&=\frac{1}{144} \sqrt{2g}\,\dee{t}\cr \int \big(h^{3/2}-12h^{1/2}\big)\dee{h}&=\int\frac{1}{144} \sqrt{2g}\,\dee{t}\cr \frac{h^{5/2}}{5/2}-12\frac{h^{3/2}}{3/2}&=\frac{1}{144} \sqrt{2g}\,t+C\end{align*}

At time 00, the height is 1212, so C=125/25/2−12123/23/2=125/2(25−23)=−415125/2C=\frac{12^{5/2}}{5/2}-12\frac{12^{3/2}}{3/2} =12^{5/2}\big(\frac{2}{5}-\frac{2}{3}\big) =-\frac{4}{15}12^{5/2}, which yields

h5/25/2−12h3/23/2=11442g t−415125/2\begin{align*}\frac{h^{5/2}}{5/2}-12\frac{h^{3/2}}{3/2}&=\frac{1}{144} \sqrt{2g}\,t -\frac{4}{15}12^{5/2}\end{align*}

We want to find the time tt when the height is h=0h=0.

0=11442g t−415125/211442g t=415125/2t=4×144151252g=38.4124416g≈2,394 sec ≈0.665 hr\begin{align*}0&=\frac{1}{144} \sqrt{2g}\,t-\frac{4}{15}12^{5/2}\\ \frac{1}{144} \sqrt{2g}\,t&=\frac{4}{15}12^{5/2}\\ t&=\frac{4\times 144}{15} \sqrt{\frac{12^5}{2g}}\\ &=38.4 \sqrt{\frac{124416}{g}}\approx 2,394\,\text{sec }\approx 0.665\, \text{hr}\end{align*}
Q34Stage 3Past exam · 2001A

Consider the equation

f(x)=3+∫0x(f(t)−1)(f(t)−2) dt\begin{align*} f(x)=3+\int_0^x\big(f(t)-1\big)\big(f(t)-2\big)\ \dee{t} \end{align*}

  1. What is f(0)f(0)?

  2. Find the differential equation satisfied by f(x)f(x).

  3. Solve the initial value problem determined in (a) and (b).

Hint

The fundamental theorem of calculus will be useful in part (b).

Answer

(a) 33 (b) y′=(y−1)(y−2)y'=(y-1)(y-2) (c) f(x)=4−ex2−exf(x)=\dfrac{4-e^x}{2-e^x}

Full solution

(a) Setting x=0x=0 gives

f(0)=3+∫00(f(t)−1)(f(t)−2) dt=3\begin{align*} f(0)=3+\int_0^0\big(f(t)-1\big)\big(f(t)-2\big)\ \dee{t}= 3 \end{align*}

(b) By the Fundamental Theorem of Calculus part 1,

f′(x)=ddx∫0x(f(t)−1)(f(t)−2) dt=(f(x)−1)(f(x)−2)\begin{align*} f'(x)=\diff{}{x}\int_0^x\big(f(t)-1\big)\big(f(t)-2\big)\ \dee{t} =\big(f(x)-1\big)\big(f(x)-2\big) \end{align*}

Thus y=f(x)y=f(x) obeys the differential equation y′=(y−1)(y−2)y'=(y-1)(y-2).

(c) If y≠1,2y\ne 1,2,

dydx=(y−1)(y−2)dy(y−1)(y−2)=dx∫dy(y−1)(y−2)=∫dx\begin{align*}\diff{y}{x}&=(y-1)(y-2) \\ \frac{\dee{y}}{(y-1)(y-2)}&=\dee{x}\\ \int\frac{\dee{y}}{(y-1)(y-2)}&=\int\dee{x}\end{align*}

Using the method of partial fractions,

∫(1y−2−1y−1)dy=∫dxlog⁡∣y−2∣−log⁡∣y−1∣=x+Clog⁡∣y−2y−1∣=x+C\begin{align*}\int\left(\frac{1}{y-2}-\frac{1}{y-1}\right)\dee{y}&=\int \dee{x} \\ \log|y-2|-\log|y-1|&=x+C\\ \log\left|\frac{y-2}{y-1}\right|&=x+C\end{align*}

Observe that dydx=(y−1)(y−2)>0\diff{y}{x}=(y-1)(y-2)>0 for all y≥2y\ge 2. That is, f(x)f(x) is increasing at all xx for which f(x)>2f(x)>2. As f(0)=3f(0)=3, f(x)f(x) increases for all x≥0x\ge 0, and f(x)≥3f(x)\ge 3 for all x≥0x\ge 0. So we may drop the absolute value signs.

log⁡f(x)−2f(x)−1=x+Cf(x)−2f(x)−1=eCex\begin{align*}\log\frac{f(x)-2}{f(x)-1}&=x+C\\ \frac{f(x)-2}{f(x)-1}&=e^Ce^x\end{align*}

At x=0x=0, f(x)−2f(x)−1=12\frac{f(x)-2}{f(x)-1}=\frac{1}{2} so eC=12e^C=\frac{1}{2}.

f(x)−2f(x)−1=12ex2f(x)−4=[f(x)−1]ex[2−ex]f(x)=4−exf(x)=4−ex2−ex\begin{align*}\frac{f(x)-2}{f(x)-1}&=\frac{1}{2} e^x\\ 2f(x)-4&=[f(x)-1]e^x\\ [2-e^x]f(x)&=4-e^x \\ f(x)&=\frac{4-e^x}{2-e^x}\end{align*}
Q35Stage 3Past exam · 2002A

A tank 2 m tall is to be made with circular cross–sections with radius r=ypr=y^p. Here yy measures the vertical distance from the bottom of the tank and pp is a positive constant to be determined. You may assume that when the tank drains, it obeys Torricelli's law, that is

A(y)dydt=−cy\begin{align*} A(y)\diff{y}{t}=-c\sqrt{y} \end{align*}

for some constant cc where A(y)A(y) is the cross–sectional area of the tank at height yy. It is desired that the tank be constructed so that the top half (y=2y=2 to y=1y=1) takes exactly the same amount of time to drain as the bottom half (y=1y=1 to y=0y=0). Determine the value of pp so that the tank has this property. Note: it is not possible or necessary to find cc for this question.

Hint

For any p>0p>0, determine first y(t)y(t) (in terms of pp and cc) and then the times (also depending on pp and cc) at which y=2y=2, y=1y=1 and y=0y=0. The condition that “the top half takes exactly the same amount of time to drain as the bottom half” then gives an equation that determines pp.

Answer

p=14p=\frac{1}{4}

Full solution

Suppose that at time tt (measured in hours starting at, say, noon), the water in the tank has height yy, which is between 0 and 2 metres. At that time, the top surface of the water forms a circular disk of radius r=ypr=y^p and area A(y)=πy2pA(y)=\pi y^{2p}. Thus, by Torricelli's law,

πy2pdydt=−cy−πc⋅y2p−12dy= dt∫−πc⋅y2p−12dy= ∫dt−πc⋅y2p+122p+12+d=t\begin{align*}\pi y^{2p}\diff{y}{t}&=-c\sqrt{y}\\ -\frac{\pi}{c}\cdot y^{2p-{1\over 2}}\dee{y}&=\,\dee{t}\\ \int-\frac{\pi}{c}\cdot y^{2p-{1\over 2}}\dee{y}&=\,\int \dee{t}\\ -\frac{\pi}{c}\cdot\frac{y^{2p+{1\over 2}}}{2p+{1\over 2}}+d &=t\end{align*}

for some constant dd. At time t=0t=0, the height is y=2y=2, so d=πc⋅22p+122p+12d=\displaystyle\frac{\pi}{c}\cdot\frac{2^{2p+{1\over 2}}}{2p+{1\over 2}} .

t=πc(22p+122p+12−y2p+122p+12)=πc(2p+12)(22p+12−y2p+12)\begin{align*}t&=\frac{\pi}{c}\bigg(\frac{2^{2p+{1\over 2}}}{2p+{1\over 2}} -\frac{y^{2p+{1\over 2}}}{2p+{1\over 2}}\bigg)\\ &=\frac{\pi}{c(2p+\frac12)}\left(2^{2p+\frac12}-y^{2p+\frac12}\right)\end{align*}

The time at which the height is 11 is obtained by subbing y=1y=1 into this formula. The time at which the height is 00 is obtained by subbing y=0y=0 into this formula. Thus the condition that the top half (y=2y=2 to y=1y=1) takes exactly the same amount of time to drain as the bottom half (y=1y=1 to y=0y=0) is:

t(2)−t(1)=t(1)−t(0)0−t(1)=t(1)−t(0)t(0)=2t(1)πc(2p+12)(22p+12−02p+12)=2πc(2p+12)(22p+12−12p+12)22p+12=2(22p+12−1)22p+12=2⋅22p+12−22=22p+121=2p+12p=14\begin{align*} t(2)-t(1)&=t(1)-t(0)\\ 0-t(1)&=t(1)-t(0)\\ t(0)&=2t(1)\\ \frac{\pi}{c(2p+\frac12)}\left(2^{2p+\frac12}-0^{2p+\frac12}\right)&= 2\frac{\pi}{c(2p+\frac12)}\left(2^{2p+\frac12}-1^{2p+\frac12}\right)\\ 2^{2p+\frac12}&= 2\left(2^{2p+\frac12}-1\right)\\ 2^{2p+\frac12}&= 2\cdot 2^{2p+\frac12}-2\\ 2&=2^{2p+\frac12}\\ 1&=2p+\frac12\\ p&=\frac14 \end{align*}
Q36Stage 3

Suppose f(t)f(t) is a continuous, differentiable function and the root mean square of f(t)f(t) on [a,x][a,x] is equal to the average of f(t)f(t) on [a,x][a,x] for all xx. That is,

1x−a∫axf(t) d(t)=1x−a∫axf2(t) dt\begin{equation} \frac{1}{x-a}\int_a^xf(t)\,\dee(t)=\sqrt{\frac{1}{x-a}\int_a^x f^2(t)\,\dee{t}}\tag{$*$} \end{equation}

You may assume x>ax>a.

  1. Guess a function f(t)f(t) for which the average of f(t)f(t) is the same as the root mean square of f(t)f(t) on any interval.

  2. Differentiate both sides of the given equation.

  3. Simplify your answer from (b) by using Equation (∗*) to replace all terms containing ∫axf2(t) dt\int_a^x f^2(t)\,\dee{t} with terms containing ∫axf(t) dt\int_a^x f(t)\,\dee{t}.

  4. Let Y(x)=∫axf(t) dtY(x) = \int_a^x f(t)\,\dee{t}, so the equation from (c) becomes a differential equation. Find all functions that satisfy it.

  5. What is f(t)f(t)?

Hint

For (a), think of a very simple function.

The equation in the question statement is equivalent to the equation

1x−a∫axf(t) d(t)=∫axf2(t) dt\frac{1}{\sqrt{x-a}}\int_a^xf(t)\,\dee(t)=\sqrt{\int_a^x f^2(t)\,\dee{t}}

which is, in some cases, easier to use.

For (d), you'll want to let Y(x)=∫axf(t) dtY(x)=\int_a^x f(t)\,\dee{t}, and use the quadratic equation.

Answer

  1. One possible answer: f(t)=0f(t)=0

  2. 1x−a[f(x)−12(x−a)∫axf(t) dt]=f2(x)2∫axf2(t) dt\displaystyle \frac{1}{\sqrt{x-a}}\left[f(x) - \frac{1}{2(x-a)}\int_a^x f(t)\,\dee{t}\right] =\frac{f^2(x)}{2\sqrt{\int_a^xf^2(t)\,\dee{t}}}

  3. 2x−a∫axf(t) dt[f(x)−12(x−a)∫axf(t) dt]=f2(x)\displaystyle \frac{2}{x-a}\int_a^x f(t)\,\dee{t}\left[f(x) - \frac{1}{2(x-a)}\int_a^x f(t)\,\dee{t}\right]=f^2(x)

  4. Y(x)=D(x−a)Y(x) = D(x-a), where DD is any constant

  5. f(t)=Df(t)=D, for any nonnegative constant DD

Full solution

  1. If we let f(t)=0f(t)=0 for all tt, then its average over any interval is 0, as is its root mean square.

  2. Let's start by simplifying the given equation.

    1x−a∫axf(t) dt=1x−a∫axf2(t) dt1x−a∫axf(t) dt=∫axf2(t) dtddx{1x−a∫axf(t) dt}=ddx{∫axf2(t) dt}\begin{align} \notag \frac{1}{x-a}\int_a^x f(t)\,\dee{t}&=\sqrt{\frac{1}{x-a}\int_a^x f^2(t)\,\dee{t}}\\ \frac{1}{\sqrt{x-a}}\int_a^x f(t)\,\dee{t}&=\sqrt{\int_a^x f^2(t)\,\dee{t}} \\ \color{red}\diff{}{x}\left\{\frac{1}{\sqrt{x-a}}\int_a^x f(t)\,\dee{t}\right\}&=\color{blue}\diff{}{x}\left\{\sqrt{\int_a^x f^2(t)\,\dee{t}}\right\} \end{align}

    For the derivative on the left, we use the product rule and the Fundamental Theorem of Calculus, part 1.

    ddx{1x−a∫axf(t) dt}=ddx{1x−a}∫axf(t) dt+1x−a⋅ddx{∫axf(t) dt}=−12x−a3∫axf(t) dt+f(x)x−a=1x−a[f(x)−12(x−a)∫axf(t) dt]\begin{align*}\color{red}\diff{}{x}\left\{\frac{1}{\sqrt{x-a}}\int_a^x f(t)\,\dee{t}\right\}&=\diff{}{x}\left\{\frac{1}{\sqrt{x-a}}\right\}\int_a^x f(t)\,\dee{t} + \frac{1}{\sqrt{x-a}}\cdot\diff{}{x}\left\{\int_a^x f(t)\,\dee{t}\right\}\\ &=-\frac{1}{2\sqrt{x-a}^3}\int_a^x f(t)\,\dee{t} + \frac{f(x)}{\sqrt{x-a}} \\ &=\color{red}\frac{1}{\sqrt{x-a}}\left[f(x) - \frac{1}{2(x-a)}\int_a^x f(t)\,\dee{t}\right]\end{align*}

    For the derivative on the right in Equation (3.2), we use the chain rule and the Fundamental Theorem of Calculus, part 1.

    ddx{∫axf2(t) dt}=12(∫axf2(t) dt)−12⋅ddx{∫axf2(t) dt}=f2(x)2∫axf2(t) dt\begin{align*}\color{blue}\diff{}{x}\left\{\sqrt{\int_a^x f^2(t)\,\dee{t}}\right\}&=\frac{1}{2}\left(\int_a^x f^2(t)\,\dee{t}\right)^{-\frac12}\cdot\diff{}{x}\left\{\int_a^x f^2(t)\,\dee{t}\right\}\\ &=\color{blue}\frac{f^2(x)}{2\sqrt{\int_a^xf^2(t)\,\dee{t}}}\end{align*}

    So, Equation (3.2) yields the following:

    1x−a[f(x)−12(x−a)∫axf(t) dt]=f2(x)2∫axf2(t) dt\begin{align} \color{red}\frac{1}{\sqrt{x-a}}\left[f(x) - \frac{1}{2(x-a)}\int_a^x f(t)\,\dee{t}\right] &=\color{blue}\frac{f^2(x)}{2\sqrt{\int_a^xf^2(t)\,\dee{t}}} \end{align}

  3. From Equation (3.1), ∫axf2(t) dt=1x−a∫axf(t) dt\sqrt{\int_a^x f^2(t)\,\dee{t}} = \frac{1}{\sqrt{x-a}}\int_a^x f(t)\,\dee{t}.

    1x−a[f(x)−12(x−a)∫axf(t) dt]=f2(x)21x−a∫axf(t) dt2x−a∫axf(t) dt[f(x)−12(x−a)∫axf(t) dt]=f2(x)\begin{align*} \frac{1}{\sqrt{x-a}}\left[f(x) - \frac{1}{2(x-a)}\int_a^x f(t)\,\dee{t}\right]&=\frac{f^2(x)}{2\frac{1}{\sqrt{x-a}}\int_a^x f(t)\,\dee{t}}\\ \frac{2}{x-a}\int_a^x f(t)\,\dee{t}\left[f(x) - \frac{1}{2(x-a)}\int_a^x f(t)\,\dee{t}\right]&=f^2(x) \end{align*}

  4. Now what we have is a differential equation, although it might not look like it. Let Y=∫axf(t) dtY = \int_a^x f(t)\,\dee{t}. Then dYdx=f(x)\diff{Y}{x} = f(x).

    2x−aY[dYdx−12(x−a)Y]=(dYdx)2\begin{align}\frac{2}{x-a}Y\left[\diff{Y}{x} - \frac{1}{2(x-a)}Y\right]&=\left(\diff{Y}{x}\right)^2\end{align}

    We're used to solving differential equations of the form dYdx=\diff{Y}{x}=(something). So, let's manipulate our equation until it has this form.

    (dYdx)2−(2Yx−a)(dYdx)+(Yx−a)2=0\begin{align}\notag \left(\diff{Y}{x}\right)^2-\left(\frac{2Y}{x-a}\right)\left(\diff{Y}{x}\right)+\left(\frac{Y}{x-a}\right)^2&=0\end{align}

    This is a quadratic equation, with variable dYdx\diff{Y}{x}. Its solutions are:

    dYdx=(2Yx−a)±(2Yx−a)2−4⋅(Yx−a)22=2Yx−a±02=Yx−a\begin{align}\notag \diff{Y}{x}&=\frac{\left(\frac{2Y}{x-a}\right)\pm\sqrt{\left(\frac{2Y}{x-a}\right)^2-4\cdot\left(\frac{Y}{x-a}\right)^2}}{2}\\ \notag&=\frac{\frac{2Y}{x-a}\pm 0}{2}\\ \notag&=\frac{Y}{x-a}\end{align}

    This gives us the separable differential equation

    dYdx=Yx−adYY=dxx−a∫dYY=∫dxx−alog⁡∣Y∣=log⁡∣x−a∣+C∣Y∣=elog⁡(x−a)+C=(x−a)eCY=D(x−a)\begin{align}\notag\diff{Y}{x}&=\frac{Y}{x-a} \\ \frac{\dee{Y}}{Y}&=\frac{\dee{x}}{x-a} \\ \notag\int\frac{\dee{Y}}{Y}&=\int\frac{\dee{x}}{x-a}\\ \notag\log|Y|&=\log|x-a|+C\\ \notag|Y|&=e^{\log(x-a)+C} = (x-a)e^C\\ \notag Y&=D(x-a)\end{align}

    where DD is some constant, eCe^C or −eC-e^C. Note this covers all real constants except D=0D=0. If D=0D=0, then Y(x)=0Y(x)=0 for all xx. This function also satisfies Equation (3.4), so indeed,

    Y(x)=D(x−a)\begin{equation} Y(x)=D(x-a) \end{equation}

    for any constant DD is the family of equations satisfying our differential equation.

    Remark: the reason we “lost" the solution Y(x)=0Y(x)=0 is that in Equation (3.5), we divided by YY, thus tacitly assuming it was not identically 0.

  5. Remember Y=∫axf(t) dtY=\int_a^x f(t)\,\dee{t}. So, Equation (3.6) tells us:

    ∫axf(t) dt=D(x−a)ddx{∫axf(t) dt}=ddx{D(x−a)}f(x)=D\begin{align*} \int_a^x f(t)\,\dee{t}&=D(x-a)\\ \diff{}{x}\left\{\int_a^x f(t)\,\dee{t}\right\}&=\diff{}{x}\{D(x-a)\}\\ f(x)&=D \end{align*}

    We should check that this function works.

    favg=1x−a∫axD dt=1x−a[Dt]t=at=x=Dx−Dax−a=DfRMS=1x−a∫axD2 dt=1x−a[D2x]t=at=x=D2x−D2ax−a=D2=∣D∣\begin{align*} f_{\text{avg}} &= \frac{1}{x-a}\int_a^x D\,\dee{t} = \frac{1}{x-a}\Big[Dt\Big]_{t=a}^{t=x} = \frac{Dx-Da}{x-a}=D\\ f_{\text{RMS}} &= \sqrt{\frac{1}{x-a}\int_a^x D^2\,\dee{t}} =\sqrt{\frac{1}{x-a}\Big[D^2x\Big]_{t=a}^{t=x}}=\sqrt{\frac{D^2x-D^2a}{x-a}}=\sqrt{D^2}=|D| \end{align*}

    So, f(x)=Df(x)=D works only if DD is nonnegative.

    That is: the only functions whose average matches their root square mean over every interval are constant, nonnegative functions.

    Remark: it was step (c) where we introduced the erroneous answer f(x)=Df(x)=D, D<0D<0 to our solution. In Equation (3.3), f(x)=Df(x)=D is not a solution if D<0D<0:

    1x−a[f(x)−12(x−a)∫axf(t) dt]=f2(x)2∫axf2(t) dt1x−a[D−12(x−a)∫axD dt]=D22∫axD2 dt1x−a[D−12(x−a)D(x−a)]=D22D2(x−a)1x−a[12D]=D22∣D∣x−aD=D2∣D∣=∣D∣\begin{align*} \frac{1}{\sqrt{x-a}}\left[f(x) - \frac{1}{2(x-a)}\int_a^x f(t)\,\dee{t}\right]&=\frac{f^2(x)}{2\sqrt{\int_a^x f^2(t)\,\dee{t}}}\\ \frac{1}{\sqrt{x-a}}\left[D - \frac{1}{2(x-a)}\int_a^x D\,\dee{t}\right]&=\frac{D^2}{2\sqrt{\int_a^x D^2\,\dee{t}}}\\ \frac{1}{\sqrt{x-a}}\left[D - \frac{1}{2(x-a)}D(x-a)\right]&=\frac{D^2}{2\sqrt{ D^2(x-a)}}\\ \frac{1}{\sqrt{x-a}}\left[\frac{1}{2}D \right]&=\frac{D^2}{2|D|\sqrt{x-a}}\\ D&=\frac{D^2}{|D|}=|D| \end{align*}

    In (c), we replace ∫axf2(t) dt\sqrt{\int_a^x f^2(t)\,\dee{t}}, which cannot be negative, with 1x−a∫axf(t) dt\frac{1}{\sqrt{x-a}}\int_a^x f(t)\,\dee{t}, which could be negative if f(t)=D<0f(t)=D<0. Indeed, if f(t)=Df(t)=D, then ∫axf2(t) dt=∣D∣x−a\sqrt{\int_a^x f^2(t)\,\dee{t}} = |D|\sqrt{x-a}, while 1x−a∫axf(t) dt=Dx−a\frac{1}{\sqrt{x-a}}\int_a^x f(t)\,\dee{t}=D\sqrt{x-a}. It is at this point that negative functions creep into our solution.

Q37Stage 3

Find the function y(x)y(x) such that

d2ydx2=2y3⋅dydx\ddiff{2}{y}{x}=\frac{2}{y^3}\cdot\diff{y}{x}

and if x=−116log⁡3x=-\frac{1}{16}\log 3, then y=1y=1 and dydx=3\diff{y}{x}=3.

You do not need to solve for yy explicitly.

Hint

Start by antidifferentiating both sides of the equation with respect to xx.

Answer

x=14(y−1+14log⁡∣2y−12y+1∣)\displaystyle x=\frac{1}{4}\left(y-1+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)

Full solution

We start by antidifferentiating both sides with respect to xx.

∫(d2ydx2) dx=∫(2y3⋅dydx) dx\begin{align*}\int \left(\ddiff{2}{y}{x}\right)\,\dee{x}&=\int\left(\frac{2}{y^3}\cdot\diff{y}{x}\right)\,\dee{x}\end{align*}

The right integral is in exactly the form we would use for a change of variables (substitution) to yy.

dydx=∫(2y3) dy=−1y2+C\begin{align*}\diff{y}{x}&=\int\left(\frac{2}{y^3}\right)\,\dee{y} = -\frac{1}{y^2}+C\end{align*}

When y=1y=1, dydx=3\diff{y}{x}=3.

3=−11+CC=4\begin{align*}3& =-\frac{1}{1}+C\\ C&=4\end{align*}

So,

dydx=−1y2+4\begin{align*}\diff{y}{x}&=-\frac{1}{y^2}+4\end{align*}

This is a separable differential equation.

dydx=4y2−1y2y24y2−1 dy=dx∫y24y2−1 dy=∫dx\begin{align*}\diff{y}{x}&=\frac{4y^2-1}{y^2} \\ \frac{y^2}{4y^2-1}\,\dee{y}&=\dee{x}\\ \int \frac{y^2}{4y^2-1}\,\dee{y}&=\int \dee{x}\tag{$*$}\end{align*}

We can evaluate the left integral with partial fractions, but because the numerator has the same degree as the denominator, we have to simplify first. We do this by inspection, but you can also use long division.

y24y2−1=14(4y2−1)+144y2−1=14(1+14y2−1)=14(1+1(2y−1)(2y+1))=14(1+1/22y−1−1/22y+1)\begin{align*} \frac{y^2}{4y^2-1}&=\frac{\frac{1}{4}(4y^2-1)+\frac{1}{4}}{4y^2-1}\\ &=\frac{1}{4}\left(1+\frac{1}{4y^2-1}\right)\\ &=\frac{1}{4}\left(1+\frac{1}{(2y-1)(2y+1)}\right)\\ &=\frac{1}{4}\left(1+\frac{1/2}{2y-1}-\frac{1/2}{2y+1}\right) \end{align*}

Now, we return to (∗*).

∫dx=∫y24y2−1 dy=∫14(1+1/22y−1−1/22y+1) dy=14(y+14log⁡∣2y−1∣−14log⁡∣2y+1∣)=14(y+14log⁡∣2y−12y+1∣)x+C=14(y+14log⁡∣2y−12y+1∣)\begin{align*}\int \dee{x}&=\int \frac{y^2}{4y^2-1}\,\dee{y}\\ &= \int \frac{1}{4}\left(1+\frac{1/2}{2y-1}-\frac{1/2}{2y+1}\right)\,\dee{y}\\ &= \frac{1}{4}\left(y+\frac{1}{4}\log|2y-1| - \frac{1}{4}\log|2y+1|\right)\\ &= \frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\\ x+C&=\frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\end{align*}

When x=−116log⁡3x=-\frac{1}{16}\log 3, y=1y=1.

−116log⁡3+C=14(1+14log⁡∣2−12+1∣)=14+116log⁡13C=14\begin{align*}-\frac{1}{16}\log 3 +C &=\frac{1}{4}\left(1+\frac14\log\left| \frac{2-1}{2+1}\right|\right) = \frac{1}{4}+\frac{1}{16}\log\frac{1}{3}\\ C&=\frac{1}{4}\end{align*}

So,

x+14=14(y+14log⁡∣2y−12y+1∣)x=14(y−1+14log⁡∣2y−12y+1∣)\begin{align*}x+\frac{1}{4}&=\frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\\ x&=\frac{1}{4}\left(y-1+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\end{align*}

We can check our answer by differentiating with respect to xx.

x=14(y−1+14log⁡∣2y−12y+1∣)4x=y−1+14log⁡∣2y−1∣−14log⁡∣2y+1∣ddx{4x}=ddx{y−1+14log⁡∣2y−1∣−14log⁡∣2y+1∣}4=dydx+14⋅2dydx2y−1−14⋅2dydx2y+14=dydx(1+1/22y−1−1/22y+1)=dydx(4y24y2−1)dydx=4y2−1y2=4−1y2\begin{align*}x&=\frac{1}{4}\left(y-1+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\\ 4x&=y-1+\frac{1}{4}\log|2y-1| -\frac{1}{4}\log|2y+1|\\ \diff{}{x}\{4x\}&=\diff{}{x}\left\{y-1+\frac{1}{4}\log|2y-1| -\frac{1}{4}\log|2y+1|\right\}\\ 4&=\diff{y}{x}+\frac{1}{4}\cdot\frac{2\diff{y}{x}}{2y-1} - \frac{1}{4}\cdot\frac{2\diff{y}{x}}{2y+1}\\ 4&=\diff{y}{x}\left(1+\frac{1/2}{2y-1} - \frac{1/2}{2y+1}\right) = \diff{y}{x}\left(\frac{4y^2}{4y^2-1}\right)\\ \diff{y}{x}&=\frac{4y^2-1}{y^2}=4-\frac{1}{y^2} \tag{$**$}\end{align*}

Differentiating with respect to xx again, using the chain rule,

d2ydx2=2y3⋅dydx\begin{align*}\ddiff{2}{y}{x}&=\frac{2}{y^3}\cdot\diff{y}{x}\end{align*}

This is exactly the differential equation we were meant to solve.

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.