Linear Ordinary Differential Equations

Definition: Linear Ordinary Differential Equations

An ordinary differential equation

$$F\left(x,y,y^{\prime },y^{\prime \prime },\dots ,y^{(n)}\right)=0$$

is linear if there exist functions $a_0,a_1,\dots ,a_n,b$ such that the ODE can be expressed as

$$a_0(x)y+a_1(x)y^{\prime }+\cdots +a_n(x)y^{(n)}=b(x)$$

First-Order Linear ODEs

A first-order linear ODE can be expressed as

$$Q(x)y^{\prime }+P(x)y=G(x)$$

for some functions $P$, $Q$ and $G$. We often need transform this into a more standard form by dividing by $Q(x)$ and imposing the condition that $Q(x)\ne 0$:

$$y^{\prime }+p(x)=g(x),$$

where $p=P/Q$ and $g=G/Q$.

Algorithm: Solving First-Order Linear ODEs

We are given the following first-order linear ODE:

$$y^{\prime }+p(x)y=g(x)$$

To solve this, we use the method of integrating factors. The goal is to find some function $\mu (x)$ such that we can use the product rule to express the left-hand side as the derivative of $(\mu (x)y)$. We can then use antidifferentiation to find the solutions.

1. Multiply both sides by a yet unknown function $\mu (x)$:

$$\mu (x)y^{\prime }+\mu (x)p(x)y=\mu (x)g(x)$$

2. In order for $\mu (x)y^{\prime }+\mu (x)p(x)y$ to be the derivative of $(\mu (x)y)$, we need $\mu (x)p(x)$ to be equal to ${\mu }^{\prime }(x)$ because the product rule gives us $(\mu (x)y{)}^{\prime }=\mu (x)y^{\prime }+{\mu }^{\prime }(x)y$:

$$\mu (x)y^{\prime }+\mu (x)p(x)y=(\mu (x)y{)}^{\prime }⟺{\mu }^{\prime }(x)=\mu (x)p(x)$$

3. Divide both sides of ${\mu }^{\prime }(x)=\mu (x)p(x)$ by $\mu (x)$, imposing the condition $\mu (x)\ne 0$:

$$\frac{1}{\mu (x)}{\mu }^{\prime }(x)=p(x)$$

4. By also imposing the condition that $\mu (x)>0$ and then antidifferentiating both sides, we can transform this further:

$$\ln (\mu (x))=\int p(x)\mathrm{dx}+C$$

5. We take the real exponential function of both sides:

$$\mu (x)={\mathrm{e}}^{\int p(x)\mathrm{dx}+C}$$

For each choice of $C\in \mathbb{R}$, the expression $\int p(x)\mathrm{dx}+C$ is some antiderivative $P(x)$ of $p(x)$. Thus, if we can find an antiderivative $P(x)$ of $p(x)$, then we can find a $\mu (x)={\mathrm{e}}^{P(x)}$ which satisfies the condition $(\mu (x)y{)}^{\prime }=\mu (x)y^{\prime }+{\mu }^{\prime }(x)y$. We can easily verify this as well by applying the chain rule and the rules for the derivative of the real exponential function:

$$(\mu (x)y{)}^{\prime }=({\mathrm{e}}^{P(x)}y{)}^{\prime }={\mathrm{e}}^{P(x)}{y}^{\prime }+({\mathrm{e}}^{P(x)}{)}^{\prime }y={\mathrm{e}}^{P(x)}{P}^{\prime }(x)y={\mathrm{e}}^{P(x)}{y}^{\prime }+{\mathrm{e}}^{P(x)}p(x)y=\mu (x)y^{\prime }+{\mu }^{\prime }(x)y$$

Moreover, since the real exponential function is always positive, we have a $\mu (x)$ which satisfies the previously imposed condition that $\mu (x)>0$.

6. We have shown how to find an appropriate $\mu (x)$, so we can proceed with the original equation:

$$(\mu (x)y{)}^{\prime }=\mu (x)g(x)$$

7. We antidifferentiate both sides:

$$\mu (x)y=\int \mu (x)g(x)\mathrm{dx}+C$$

8. To obtain the solutions, we divide both sides by $\mu (x)$:

$$y=\frac{1}{\mu (x)}\left(\int \mu (x)g(x)\mathrm{dx}+C\right)$$

For each choice of $C\in \mathbb{R}$, the expression $\int \mu (x)g(x)\mathrm{dx}+C$ is just some antiderivative $\mathcal{F}$ of $\mu (x)g(x)$. Therefore, the solutions of the original equation on some open subset $U\subseteq \mathbb{R}$ are the functions $y:U\to \mathbb{R}$ which for all $x\in U$ can be expressed for as

$$y(x)=\frac{1}{\mu (x)}\mathcal{F}(x)=\frac{1}{{\mathrm{e}}^{P(x)}}\mathcal{F}(x)={\mathrm{e}}^{-P(x)}\mathcal{F}(x),$$

where $P(x)$ is any antiderivative of $p(x)$ and $\mathcal{F}(x)$ is any antiderivative of $\mu (x)g(x)={\mathrm{e}}^{P(x)}g(x)$.

Summary

Given some open subset $U\subseteq \mathbb{R}$, the solutions of the ODE on $U$ are the functions $y:U\to \mathbb{R}$ which on $U$ can be expressed as

$$y(x)={\mathrm{e}}^{-P(x)}\mathcal{F}(x)$$

for some antiderivative $P$ of $p$ and some antiderivative $\mathcal{F}$ of ${\mathrm{e}}^{P(x)}g(x)$.

Example: $y^{\prime }-2y=4-x$

Here we have $p(x)=-2$ and $g(x)=4-x$. We want to rewrite this as

$$(\mu (x)y{)}^{\prime }=\mu (x)(4-x)$$

The antiderivatives of $p$ are given by

$$\int p(x)\mathrm{dx}=\int -2\mathrm{dx}=-2x+C$$

We choose $C=0$ to make everything simple and so

$$\mu (x)={\mathrm{e}}^{-2x}$$

The equation thus becomes

$$({\mathrm{e}}^{-2x}y{)}^{\prime }={\mathrm{e}}^{-2x}(4-x)$$

We antidifferentiate both sides:

$${\mathrm{e}}^{-2x}y=\int {\mathrm{e}}^{-2x}(4-x)\mathrm{dx}$$

Use integration by parts on the right-hand side:

$$\begin{array}{rl}\int {\mathrm{e}}^{-2x}(4-x)\mathrm{dx}& =(4-x)\left(-\frac{1}{2}{\mathrm{e}}^{2x}\right)-\int \frac{1}{2}{\mathrm{e}}^{-2x}\mathrm{dx}\\ & =-\frac{1}{2}(4-x)\left({\mathrm{e}}^{-2x}\right)-\frac{1}{2}{\mathrm{e}}^{-2x}\mathrm{dx}\\ & =-\frac{1}{2}(4-x)\left({\mathrm{e}}^{-2x}\right)-\frac{1}{2}\left(-\frac{1}{2}{\mathrm{e}}^{-2x}+C\right)\\ & =-\frac{1}{2}(4-x)\left({\mathrm{e}}^{-2x}\right)+\frac{1}{4}{\mathrm{e}}^{-2x}+C\\ & =-2{\mathrm{e}}^{-2x}+\frac{1}{2}x{\mathrm{e}}^{-2x}+\frac{1}{4}{\mathrm{e}}^{-2x}+C\\ & =\left(-2+\frac{1}{4}\right){\mathrm{e}}^{-2x}+\frac{1}{2}x{\mathrm{e}}^{-2x}+C\\ & =\left(-\frac{8}{4}+\frac{1}{4}\right){\mathrm{e}}^{-2x}+\frac{1}{2}x{\mathrm{e}}^{-2x}+C\\ & =-\frac{7}{4}{\mathrm{e}}^{-2x}+\frac{1}{2}x{\mathrm{e}}^{-2x}+C\\ & ={\mathrm{e}}^{-2x}\left(\frac{1}{2}x-\frac{7}{4}\right)+C\end{array}$$

Therefore, we have

$${\mathrm{e}}^{-2x}y={\mathrm{e}}^{-2x}\left(\frac{1}{2}x-\frac{7}{4}\right)+C$$

and so the solutions are

$$y(x)=C{\mathrm{e}}^{2x}+\frac{1}{2}x-\frac{7}{4}$$

Example: $x{y}^{\prime }+2y=4x^2$

TODO