2.4 Separable Differential Equations

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

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

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

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

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

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

false

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

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

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

Using the definition of a derivative,

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

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

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

$\diff{p}{t}=\alpha p(t)\big(1-p(t)\big)$, for some constant $\alpha$.

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

(a) $-1$ (b) $0$ (c) $0.5$
(d) Two possible answers are shown below:

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

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 $y$ increases) to approximate most of the points.

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

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

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

$y = \log(x^2+2)$

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

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

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

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

Simplify the equation.

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

Be careful with the arbitrary constant.

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

Start by cross-multiplying.

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

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

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

Be careful about signs.

$y = \sqrt{10x^3 + 4x^2 + 6x - 4}$

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

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

Move the $y$ 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 $y=-1$ when $x=1$. This suggests how to deal with absolute values.

$y=\frac{1}{1-2x}$

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

$f(x) = e\cdot e^{x^2/2}$

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

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

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

$\displaystyle\diff{}{x}\{\sec x\} = \sec x \tan x$

$\displaystyle y^2+\frac{2}{3}(y^2-4)^{3/2}=2\sec x +2$

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

$12\text{ weeks}$

$t=\displaystyle\sqrt{\frac{m}{kg}}\arctan \left(\sqrt{\frac{k}{mg}}\,v_0\right)$

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

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

The method of partial fractions will help you integrate.

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

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.

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

Be careful about signs.

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

(a) $P=\dfrac{4}{1+e^{-4t}}$ (b) At $t=\dfrac{1}{2}$, $P\approx 3.523$. As $t\rightarrow\infty$, $P\rightarrow 4$.

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.

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

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.

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

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

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

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

$y(x)=\dfrac{4-e^{1-\cos x}}{2-e^{1-\cos x}}$. The largest allowed interval is

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

Suppose that in a very short time interval $\dee{t}$, the height of water in the tank changes by $\dee{h}$ (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.

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

Sketch the mercury in the tank at time $t$, when it has height $h$, and also at time $t+\dee{t}$, when it has height $h+\dee{h}$ (with $\dee{h}<0$). The difference between those two volumes is the volume of (essentially) a disk of thickness $-\dee{h}$. 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 $h(t)$.

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

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

(a) $3$ (b) $y'=(y-1)(y-2)$ (c) $f(x)=\dfrac{4-e^x}{2-e^x}$

For any $p>0$, determine first $y(t)$ (in terms of $p$ and $c$) and then the times (also depending on $p$ and $c$) at which $y=2$, $y=1$ and $y=0$. 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 $p$.

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

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

The equation in the question statement is equivalent to the equation

which is, in some cases, easier to use.

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