3.3: Linear Equations

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The linear first-order differential equation (linear in $y$ and its derivative) can be written in the form $$\label{eq:1}\frac{dy}{dx}+p(x)y=g(x),$$ with the initial condition $y(x_0) = y_0$. Linear first-order equations can be integrated using an integrating factor $\mu(x)$. We multiply $\eqref{eq:1}$ by $\mu(x)$, $$\label{eq:2}\mu (x)\left[\frac{dy}{dx}+p(x)y\right]=\mu (x)g(x),$$ and try to determine $\mu (x)$ so that $$\label{eq:3}\mu (x)\left[\frac{dy}{dx}+p(x)y\right]=\frac{d}{dx}[\mu (x)y].$$

Figure $\PageIndex{1}$: Solution of the following ODE: $(3+2y)y'=2\cos 2x$, $y(0)=-1$.

Equation $\eqref{eq:2}$ then becomes $$\label{eq:4}\frac{d}{dx}[\mu (x)y]=\mu (x)g(x).$$

Equation $\eqref{eq:4}$ is easily integrated using $\mu (x_0)=\mu_0$ and $y(x_0)=y_0$:

$$\mu (x)y-\mu_0 y_0=\int_{x_0}^x\mu (x)g(x)dx,\nonumber$$ or $$\label{eq:5}y=\frac{1}{\mu (x)}\left(\mu_0y_0+\int_{x_0}^x\mu(x)g(x)dx\right).$$

It remains to determine $\mu(x)$ from $\eqref{eq:3}$. Differentiating and expanding $\eqref{eq:3}$ yields $$\mu\frac{dy}{dx}+p\mu y=\frac{d\mu}{dx}y+\mu\frac{dy}{dx};\nonumber$$ and upon simplifying, $$\label{eq:6}\frac{d\mu}{dx}=p\mu.$$

Equation $\eqref{eq:6}$ is separable and can be integrated:

$$\begin{aligned}\int_{\mu_0}^{\mu}\frac{d\mu}{\mu}&=\int_{x_0}^xp(x)dx, \\ \ln\frac{\mu}{\mu_0}&=\int_{x_0}^xp(x)dx, \\ \mu(x)&=\mu_0\exp\left(\int_{x_0}^xp(x)dx\right).\end{aligned}$$

Notice that since $\mu_0$ cancels out of $\eqref{eq:5}$, it is customary to assign $\mu_0 = 1$. The solution to $\eqref{eq:1}$ satisfying the initial condition $y(x_0) = y_0$ is then commonly written as $$y=\frac{1}{\mu(x)}\left(y_0+\int_{x_0}^x\mu(x)g(x)dx\right),\nonumber$$ with $$\mu(x)=\exp\left(\int_{x_0}^xp(x)dx\right)\nonumber$$ the integrating factor. This important result finds frequent use in applied mathematics.

Example $\PageIndex{1}$

Solve $\frac{dy}{dx}+2y=e^{-x}$, with $y(0)=3/4$.

Solution

Note that this equation is not separable. With $p(x) = 2$ and $g(x) = e^{−x}$, we have $$\begin{aligned}\mu(x)&=\exp\left(\int_0^x 2dx\right) \\ &=e^{2x},\end{aligned}$$ and $$\begin{aligned}y&=e^{-2x}\left(\frac{3}{4}+\int_0^x e^{2x}e^{-x}dx\right) \\ &=e^{-2x}\left(\frac{3}{4}+\int_0^x e^xdx\right) \\ &=e^{-2x}\left(\frac{3}{4}+(e^x-1)\right) \\ &=e^{-2x}\left(e^x-\frac{1}{4}\right) \\ &=e^{-x}\left(1-\frac{1}{4}e^{-x}\right).\end{aligned}$$

Example $\PageIndex{2}$

Solve $\frac{dy}{dx}-2xy=x$, with $y(0)=0$.

Solution

This equation is separable, and we solve it in two ways. First, using an integrating factor with $p(x) = −2x$ and $g(x) = x$:

$$\begin{aligned}\mu(x)&=\exp\left(-2\int_0^x xdx\right) \\ &=e^{-x^2},\end{aligned}$$ and $$y=e^{x^2}\int_0^x xe^{-x^2}dx.\nonumber$$

The integral can be done by substitution with $u = x^2$, $du = 2xdx$:

$$\begin{aligned}\int_0^x xe^{-x^2}dx&=\frac{1}{2}\int_0^{x^2}e^{-u}du \\ &=-\frac{1}{2}e^{-u}]_0^{x^2}\\ &=\frac{1}{2}\left(1-e^{-x^2}\right).\end{aligned}$$

Therefore, $$\begin{aligned}y&=\frac{1}{2}e^{x^2}\left(1-e^{-x^2}\right) \\ &=\frac{1}{2}\left(e^{x^2}-1\right).\end{aligned}$$

Second, we integrate by separating variables:

$$\begin{aligned}\frac{dy}{dx}-2xy&=x, \\ \frac{dy}{dx}&=x(1+2y), \\ \int_0^y\frac{dy}{1+2y}&=\int_0^x xdx, \\ \frac{1}{2}\ln (1+2y)&=\frac{1}{2}x^2, \\ 1+2y&=e^{x^2}, \\ y&=\frac{1}{2}\left(e^{x^2}-1\right).\end{aligned}$$

The results from the two different solution methods are the same, and the choice of method is a personal preference.