Linear differential equations: Introduction

Let $q(x),p_0(x),p_1(x),\cdots ,p_{n-1}(x)$ be functions of $x$. The equation

$$\begin{array}{}\text{(Eq:linode)}& y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots +p_1(x)y^{\prime }+p_0(x)y+q(x)=0\end{array}$$

of an unknown function $y=y(x)$ is called an $n$-th order linear differential equation. If $q(x)=0$, then (Eq:linode) is said to be homogeneous; otherwise, it is said to be inhomogeneous.

Let's rewrite (Eq:linode) using differential operators. Let $D=\frac{d}{dx}$ denote the differential operator with respect to $x$. That is, $Dy=\frac{d}{dx}y=y^{\prime }$ and $D^{n}y=\frac{d^n}{d{x}^n}y=y^{(n)}$, etc. By combining these operators, we can define a new operator $E$ by

$$\begin{array}{}\text{(eq:diffop)}& E=D^n+p_{n-1}(x)D^{n-1}+\cdots +p_1(x)D+p_0(x).\end{array}$$

Then, (Eq:linode) is concisely denoted as

$$Ey+q(x)=0.$$

Theorem (Linear combinations of solutions of a homogeneous linear ODE)

Let $y_1(x)$ and $y_2(x)$ be functions. For any $a,b\in \mathbb{R}$, the following holds:

$$E(a{y}_1+b{y}_2)=aE{y}_1+bE{y}_2$$

where $E$ is the differential operator defined by (Eq:diffop). In particular, if $y=y_1(x)$ and $y=y_2(x)$ are the solutions of the homogeneous linear differential equation $Ey=0$, then $a{y}_1(x)+b{y}_2(x)$ is also a solution.

Proof. Let $y=a{y}_1+b{y}_2$. For $k\ge 0$, we have

$$D^{k}y=y^{(k)}=a{y}_1^{(k)}+b{y}_2^{(k)}.$$

Thus,

$$\begin{array}{rcl}Ey& =& D^{n}y+p_{n-1}(x)D^{n-1}y+\cdots +p_1(x)Dy+p_0(x)y\\ & =& a\{D^n{y}_1+p_{n-1}(x)D^{n-1}{y}_1+\cdots +p_1(x)D{y}_1+p_0(x)y_1\}\\ & & +b\{D^n{y}_2+p_{n-1}(x)D^{n-1}{y}_2+\cdots +p_1(x)D{y}_2+p_0(x)y_2\}\\ & =& aE{y}_1+bE{y}_2.\end{array}$$

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This theorem indicates two things:

1. The operator $E$ is linear.
2. The solutions of $Ey=0$ are closed under scalar (real numbers) multiplication and addition. In other words, the set of solutions of a linear ODE comprises a vector space. (This means that each solution can be regarded as a vector.)

Example. Consider 

$$y^{″}+{\omega }^{2}y=0.$$

$y_1=\cos (\omega x)$ and $y_2=\sin (\omega x)$ are solutions of this ODE. In fact, $y_1^{″}=-{\omega }^2\cos (\omega x)=-{\omega }^2{y}_1$ and $y_2^{″}=-{\omega }^2\sin (\omega x)=-{\omega }^2{y}_2$. For any $a,b\in \mathbb{R}$, we have

$$(a\cos (\omega x)+b\sin (\omega x){)}^{″}=-{\omega }^2(a\cos (\omega x)+b\sin (\omega x)).$$

Thus, $y=a{y}_1+b{y}_2$ is also a solution. □

Theorem (General solution of an inhomogeneous linear ODE)

Consider an inhomogeneous differential equation:

$$Ey+q(x)=0$$

and the corresponding homogeneous differential equation

$$Ey=0.$$

Let $y=y_0(x)$ be a special solution of $Ey+q(x)=0$. Then, the general solution of $Ey+q(x)=0$ is given by

$$y=y_0(x)+Y(x)$$

where $Y(x)$ is the general solution of $Ey=0$.

Proof. Let $y=y_0(x)+Y(x)$. Then, since $E{y}_0(x)+q(x)=0$ and $EY(x)=0$, 

$$\begin{array}{rcl}Ey+q(x)& =& \{E{y}_0(x)+EY(x)\}+q(x)\\ & =& \{E{y}_0(x)+q(x)\}+EY(x)=0.\end{array}$$

Suppose that $f(x)$ is an arbitrary solution of $Ey+q(x)=0$. Let $Y(x)=f(x)-y_0(x)$. Then, $f(x)=y_0(x)+Y(x)$. But

$$EY=Ey-E{y}_0=-q(x)-(-q(x))=0.$$

Thus, $y=Y(x)$ is a solution of $Ey=0$. ■

Example. Consider the inhomogeneous linear ODE

$$y^{\prime }+2y-3e^{4x}=0.$$

$y_0=\frac{1}{2}{e}^{4x}$ is a special solution. In fact,

$${\left(\frac{1}{2}{e}^{4x}\right)}^{\prime }+2\left(\frac{1}{2}{e}^{4x}\right)-3e^{4x}=0.$$

The corresponding homogeneous ODE

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

has a solution $Y=e^{-2x}$. In fact, 

$$(e^{-2x}{)}^{\prime }+2e^{-2x}=0.$$

Thus, the general solution is

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

where $C$ is a constant. □