Solution of Ordinary Differential Equations

Definition 10.9 (Solution of a DE). A solution of a differential equation is an expression for the dependent variable in terms of the independent variable(s) that satisfies the equation when substituted in.

Caution.

1. A differential equation need not have any solution at all — e.g. $\left(y'(x)\right)^2+(y(x))^2+1=0$ has no real solution, since $\left(y'(x)\right)^2=-\left[(y(x))^2+1\right]$ can never be non-negative.
2. Even when a solution exists, it need not be unique — for example, $y=e^{2x}$, $y=2e^{2x}$, and $y=8e^{2x}$ are all solutions of $\dfrac{dy}{dx}-2y=0$; in fact every function $y=ce^{2x}$ (any constant $c\in\mathbb R$) is a solution. Definition 10.10 (General solution). The solution containing as many arbitrary constants as the order of the differential equation is the general solution; it captures every possible solution (arbitrary constants for an ODE, arbitrary functions for a PDE). Definition 10.11 (Particular solution). Assigning particular numerical values to the arbitrary constants of a general solution — usually determined from extra given conditions — gives a particular solution. Geometrically, the general solution of a first-order equation $y'=f(x,y)$ represents a one-parameter family of curves in the $xy$-plane; e.g. $y=ce^{2x}$ is the general solution of $\dfrac{dy}{dx}-2y=0$, while $y=a\cos x+b\sin x$ (two arbitrary constants) is the general solution of $\dfrac{d^2y}{dx^2}+y=0$ — setting $a=1,b=0$ there gives the particular solution $y=\cos x$. Verification technique (worked examples). To check that a given expression solves a stated differential equation, differentiate it the required number of times and substitute back until both sides agree:

• $x^2+y^2=r^2$ (one constant $r$) differentiates to $2x+2y\dfrac{dy}{dx}=0\Rightarrow\dfrac{dy}{dx}=-\dfrac{x}{y}$ — confirming it solves $\dfrac{dy}{dx}=-\dfrac{x}{y}$.
• $y=mx+\dfrac{7}{m}$ ($m\ne0$, one constant) differentiates to $y'=m$; substituting $y'=m$ and $y$ back into $xy'+\dfrac{7}{y'}-y=0$ collapses both sides to $0$.
• $y=(x-2)+Ce^{-x^3}$ (one constant $C$) differentiates to $\dfrac{dy}{dx}=1-3x^2Ce^{-x^3}$; substituting into $\dfrac{dy}{dx}+3x^2y-4x^3=0$ again collapses both sides to $0$ after simplification. …