First-order linear differential equations

In this post, we'll see how we solve first-order linear differential equations.

Consider the following first-order homogeneous linear differential equation

$$\begin{array}{}\text{(Eq:homdiff)}& y^{\prime }+p(x)y=0.\end{array}$$ 

By separating variables, we have

$$\frac{dy}{y}=-p(x)dx.$$

Integrating both sides gives

$$\log |y|=-\int p(x)dx+c$$

so that

$$\begin{array}{}\text{(Eq:homsol)}& y=C{e}^{-\int p(x)dx}\end{array}$$

where $C$ is a constant.

Example. Let's solve

$$y^{\prime }+2y=0.$$

By separating variables, we have

$$\frac{dy}{y}=-2dx.$$

Integrating both sides,

$$\log |y|=-2x+c.$$

Exponentiating both sides, we have

$$y=C{e}^{-2x}.$$

where $C$ is a constant. □

Method of variation of parameters

Next, consider the inhomogeneous differential equation

$$\begin{array}{}\text{(Eq:inhomdiff)}& y^{\prime }+p(x)y+q(x)=0.\end{array}$$

As we have learned in a previous post, we need to find one special solution to construct the general solution. How do we find a special solution?

See also: Linear differential equations: Introduction

Here's one way. This is called the method of variation of parameters. That is, we replace the constant (paramater) $C$ in the solution (Eq:homsol) of the homogeneous case (Eq:homdiff) with a function $C(x)$:

$$y=C(x)Y(x)$$

where $Y(x)=\exp (-\int p(x)dx)$ is a solution of the homogeneous equation. Noting that $Y^{\prime }+p(x)Y=0$,

$$\begin{array}{rcl}{y}^{\prime }+p(x)y+q(x)& =& \{C(x)Y(x){\}}^{\prime }+p(x)C(x)Y(x)+q(x)\\ & =& Y(x)C^{\prime }(x)+q(x).\end{array}$$

If $y=C(x)Y(x)$ is a solution, then we must have $Y(x)C^{\prime }(x)+q(x)=0$, which gives

$$C^{\prime }(x)=-q(x)e^{\int p(x)dx}.$$

Integrating this yields

$$C(x)=-\int q(x)e^{\int p(x)dx}dx.$$

Therefore,

$$y=-e^{-\int p(x)dx}\int q(x)e^{\int p(x)dx}dx$$

is a special solution.

Example. Let's solve

$$\begin{array}{}\text{(Eq:eg2)}& y^{\prime }+2y-3e^{4x}=0.\end{array}$$

First, solving the homogeneous ODE

$$y^{\prime }+2y=0,$$

we have

$$y=C{e}^{-2x}.$$

Replacing the constant $C$ with a function $p(x)$, let $y(x)=p(x)e^{-2x}$.

Then, we have $y^{\prime }=p^{\prime }(x)e^{-2x}-2p(x)e^{-2x}$. Substituting this into (Eq:eg2),

$$p^{\prime }(x)e^{-2x}-2p(x)e^{-2x}+2p(x)e^{-2x}-3e^{4x}=0$$

so that

$$p^{\prime }(x)=3e^{6x}.$$

Integrating both sides, we have

$$p(x)=\frac{1}{2}{e}^{6x}+C$$

where $C$ is a constant. Therefore, the general solution is given by

$$y=(\frac{1}{2}{e}^{6x}+C)e^{-2x}=\frac{1}{2}{e}^{4x}+C{e}^{-2x}$$

where $C$ is a constant. □