Introduction to Differential Equations

Every topic so far has asked the forward question: given a function, what is its rate of change? A differential equation asks the reverse and far harder question: given information about the rate of change, what is the function? Nearly every quantitative law in physics, chemistry, biology and finance is written as a differential equation, and this topic covers the vocabulary (order, solution, general solution, initial value problem), the two families you can solve by hand in a first course (separable and first-order linear), and the two models they generate most often (exponential growth/decay and Newton's law of cooling).

What a differential equation is, and its order

Definition. An ordinary differential equation (ODE) is an equation relating an unknown function $y$ of a single variable $x$ to one or more of its derivatives. ("Ordinary" means one independent variable; equations involving partial derivatives in several variables are partial differential equations and are not treated here.)

Definition (order). The order of a differential equation is the order of the highest derivative that actually appears in it. Order is about which derivative, never about what power it is raised to.

Definition (first-order linear). A first-order equation is linear if it can be written as $y'+p(x)y=q(x)$: $y$ and $y'$ occur to the first power only, multiplied by functions of $x$ alone, and never inside another function. So $y'+y^{2}=x$ and $y'+\sin y=0$ are not linear.

Solutions, general solutions, particular solutions

Definition (solution). A function $\varphi$ is a solution of the equation on an interval $I$ if $\varphi$ has as many derivatives on $I$ as the equation requires, and substituting $y=\varphi(x)$ makes the equation a true statement for every $x$ in $I$.

The words "on an interval" are part of the definition, not decoration. The function $y=\dfrac{1}{x-1}$ satisfies $y'=-y^{2}$ (indeed $y'=-(x-1)^{-2}=-y^{2}$), but only on $(-\infty,1)$ or on $(1,\infty)$ — never on an interval containing $1$, where it is not even defined.

Verifying a proposed solution never requires solving anything: differentiate the candidate as many times as needed, substitute into both sides, and simplify until the two sides are visibly identical. If they are not identical for all $x$ in the interval, it is not a solution.

General vs. particular vs. singular. An $n$-th order equation typically has a family of solutions containing $n$ arbitrary constants; this family is called the general solution. Fixing the constants gives a particular solution. A solution that cannot be obtained from the family for any value of the constants is a singular solution — separable equations produce these routinely (see below).

Initial value problems

An initial value problem (IVP) is a differential equation together with conditions prescribing $y$ (and, for order $n\ge 2$, its first $n-1$ derivatives) at a single point $x_{0}$. The procedure is always: find the general solution first, then substitute the initial data to determine the constants. Imposing the condition before the constant of integration has appeared destroys the whole family.

Theorem (existence and uniqueness, first order). Consider $y'=f(x,y)$ with $y(x_{0})=y_{0}$. If $f$ and its partial derivative $f_{y}$ (with respect to $y$) are continuous on an open rectangle containing the point $(x_{0},y_{0})$, then there is an open interval about $x_{0}$ on which the problem has exactly one solution.

Both hypotheses matter, and the guarantee is only local. Continuity of $f$ alone gives existence but not uniqueness: $y'=3y^{2/3}$, $y(0)=0$ is solved both by $y\equiv 0$ and by $y=x^{3}$, and here $f_{y}=2y^{-1/3}$ is undefined at $y=0$.

Theorem (linear equations, stronger). If $p$ and $q$ are continuous on an open interval $I$ containing $x_{0}$, then $y'+p(x)y=q(x)$, $y(x_{0})=y_{0}$ has a unique solution defined on all of $I$ — not merely near $x_{0}$.

Separable equations

Form. $\dfrac{dy}{dx}=g(x)h(y)$.

Method (all four steps are compulsory).

1. Find the equilibrium solutions first. Every root $r$ of $h(y)=0$ gives the constant solution $y\equiv r$: then $\dfrac{dy}{dx}=0$ and $g(x)h(r)=0$, so both sides vanish.
2. On any interval where $h(y)\ne 0$, divide and integrate: $\int\frac{dy}{h(y)}=\int g(x)\,dx$. One constant $C$, on the right-hand side, is enough.
3. Solve for $y$ if you can; otherwise leave the answer as an implicit relation.
4. Check whether the equilibria are recovered by some value of the constant. Some are absorbed into the family; others are genuinely lost and must be listed separately as singular solutions.

Step 1 is the step everyone skips. Dividing by $h(y)$ is illegal exactly where $h(y)=0$, so those solutions can never emerge from step 2.

Why the method is legitimate (it is not "multiplying by $dx$"). Suppose $h(y(x))\ne 0$, so $\dfrac{y'}{h(y)}=g(x)$. Let $H$ be an antiderivative of $\dfrac{1}{h}$. By the chain rule $\dfrac{d}{dx}\left[H(y(x))\right]=H'(y)\,y'=\dfrac{y'}{h(y)}=g(x)$, so $H(y(x))$ and $\int g(x)\,dx$ are two antiderivatives of the same function on an interval and therefore differ by a constant.

The absolute value and the sign-absorbing constant. Integrating $\dfrac{1}{y-3}$ gives $\ln$ of the absolute value of $y-3$, so $\ln|y-3|=x^{2}+C$ becomes $|y-3|=e^{C}e^{x^{2}}$ and then $y-3=Ae^{x^{2}}$ where $A=\pm e^{C}$ is any nonzero real. The sign cannot flip partway along, because $y-3$ is continuous and never zero on the interval. Allowing $A=0$ restores the equilibrium $y\equiv 3$, so here the general solution is $y=3+Ae^{x^{2}}$ with $A$ arbitrary.

First-order linear equations: the integrating factor

Standard form (mandatory). $y'+p(x)y=q(x)$. If the equation arrives as $a(x)y'+b(x)y=c(x)$, divide through by $a(x)$ before reading off $p$, and exclude the zeros of $a$ from the interval.

Derivation. Look for a nonzero function $\mu(x)$ that turns the left side into an exact derivative:

$\mu(x)\left[y'+p(x)y\right]=\mu(x)y'+\mu(x)p(x)y \qquad\text{versus}\qquad \left(\mu y\right)'=\mu y'+\mu' y .$

These agree for every $y$ precisely when $\mu'=p\mu$. That little equation is itself separable: $\dfrac{d\mu}{\mu}=p\,dx$, so $\ln|\mu|=\int p\,dx$ and we may take

$\mu(x)=e^{\int p(x)\,dx}>0 .$

Any single antiderivative of $p$ will do — an extra $+C$ contributes a factor $e^{C}$ that cancels from both sides. With this $\mu$,

$\left(\mu y\right)'=\mu q \quad\implies\quad \mu y=\int \mu(x)q(x)\,dx+C \quad\implies\quad y=\frac{1}{\mu(x)}\left[\int \mu(x)q(x)\,dx+C\right].$

The $C$ lives inside the bracket: the term $C/\mu=Ce^{-\int p\,dx}$ is exactly the general solution of the associated homogeneous equation $y'+py=0$, and dropping it loses half the answer. Note $\mu>0$ always, so dividing by $\mu$ is never the illegal step that division by $h(y)$ was.

Exponential growth and decay

Theorem. Let $k$ be constant and let $I$ be an interval. Then $y'(t)=ky(t)$ on $I$ if and only if $y(t)=Ce^{kt}$ for some constant $C$.

Proof of the harder direction. Put $g(t)=y(t)e^{-kt}$. Then $g'=y'e^{-kt}-ky e^{-kt}=e^{-kt}\left(y'-ky\right)=0$ on $I$, and a function with zero derivative on an interval is constant there. (That last step is the Mean Value Theorem corollary: $g$ is differentiable on $I$, hence continuous on every closed subinterval and differentiable on its interior, so $g(t_{2})-g(t_{1})=g'(c)(t_{2}-t_{1})=0$. The hypothesis that the domain is an interval is what makes it work.) Hence $y(t)e^{-kt}=C$.

If $0\in I$ then $C=y(0)$. Growth when $k>0$, decay when $k<0$.

• Doubling time ($k>0$): $e^{kT}=2$, so $T=\dfrac{\ln 2}{k}$.
• Half-life ($k<0$): $e^{kT}=\dfrac12$, so $T=-\dfrac{\ln 2}{k}=\dfrac{\ln 2}{|k|}$, equivalently $k=-\dfrac{\ln 2}{T}$ and $y(t)=y_{0}2^{-t/T}$.

Newton's law of cooling

An object at temperature $T(t)$ sitting in surroundings held at the constant temperature $T_{s}$ changes temperature at a rate proportional to the temperature difference:

$\frac{dT}{dt}=k\left(T-T_{s}\right),\qquad k<0 .$

Substituting $u=T-T_{s}$ gives $u'=T'=ku$, so $u=u_{0}e^{kt}$ and

$T(t)=T_{s}+\left(T(0)-T_{s}\right)e^{kt}.$

Sanity checks built into the formula: $t=0$ returns $T(0)$; $t\to\infty$ gives $T\to T_{s}$ because $k<0$; and $T\equiv T_{s}$ is the equilibrium solution, recovered when $T(0)=T_{s}$. The same formula models heating when $T(0)<T_{s}$ — nothing changes.

Worked example 1 (separable IVP, with a lost solution)

Solve $\dfrac{dy}{dx}=-2xy^{2}$ with $y(0)=1$, and state where the solution is valid.

Step 1 — equilibria. Here $h(y)=y^{2}$, and $h(y)=0$ only at $y=0$, so $y\equiv 0$ is a solution. It fails $y(0)=1$, so it is not our answer — but keep it in view.

Step 2 — separate and integrate (valid where $y\ne 0$):

$\begin{aligned} y^{-2}\,dy &= -2x\,dx \\ \int y^{-2}\,dy &= \int -2x\,dx \\ -\frac{1}{y} &= -x^{2}+C . \end{aligned}$

Step 3 — solve for $y$. Multiply by $-1$: $\dfrac{1}{y}=x^{2}-C$. Writing $D=-C$ (still arbitrary),

$y=\frac{1}{x^{2}+D}.$

Step 4 — apply the initial condition. $1=y(0)=\dfrac{1}{D}$, so $D=1$ and

$y=\frac{1}{x^{2}+1}.$

Check. $y'=-\dfrac{2x}{\left(x^{2}+1\right)^{2}}$ and $-2xy^{2}=-\dfrac{2x}{\left(x^{2}+1\right)^{2}}$: identical. Also $y(0)=1$. Since $x^{2}+1$ is never $0$, this solution is valid for all real $x$.

Two remarks. (i) $y\equiv 0$ is a singular solution: no value of $D$ produces the zero function. (ii) The interval of validity depends on the data, not just on the equation: with $y(0)=-1$ instead we get $D=-1$ and $y=\dfrac{1}{x^{2}-1}$, valid only on $(-1,1)$, the largest interval around $x=0$ avoiding $x=\pm 1$.

Worked example 2 (first-order linear IVP)

Solve $xy'+2y=4x^{2}$ with $y(1)=2$, for $x>0$.

Standard form. Divide by $x$ (legal since $x>0$):

$y'+\frac{2}{x}\,y=4x,\qquad p(x)=\frac{2}{x},\quad q(x)=4x .$

Integrating factor. $\int p\,dx=\int\frac{2}{x}\,dx=2\ln x=\ln x^{2}$ for $x>0$, so $\mu=e^{\ln x^{2}}=x^{2}$.

Multiply and recognise the product rule.

$\begin{aligned} x^{2}y'+2xy &= 4x^{3} \\ \left(x^{2}y\right)' &= 4x^{3} \\ x^{2}y &= x^{4}+C \\ y &= x^{2}+\frac{C}{x^{2}} . \end{aligned}$

Initial condition. $y(1)=1+C=2$, so $C=1$ and

$y=x^{2}+\frac{1}{x^{2}},\qquad x>0 .$

Check. $y'=2x-\dfrac{2}{x^{3}}$, so $xy'+2y=2x^{2}-\dfrac{2}{x^{2}}+2x^{2}+\dfrac{2}{x^{2}}=4x^{2}$, and $y(1)=1+1=2$. Since $p$ and $q$ are continuous on $(0,\infty)$, the linear theorem promises this is the unique solution on the whole of $(0,\infty)$, which matches the formula's domain.

Common mistakes