Special solution of inhomogeneous linear differential equations

The general solution of an inhomogeneous linear differential equation can be obtained as the sum of a special solution and the general solution of the corresponding homogeneous differential equation. We study the method of variation of parameters in particular.

Consider the linear differential equation of the form

$$F(D)y=q(x)$$

where $F(t)$ is a polynomial and $q(x)$ is a function. When $q(x)=0$, this is a homogeneous linear differential equation. Here, we assume $q(x)\ne 0$.

Suppose we can factorize $F(t)$ as $F(t)=G(t)H(t)$. If $y=y(x)$ is a solution of $F(D)y=q$, then $z=H(D)y$ is a solution of $G(D)z=q$. Thus, the given differential equation $F(D)y=q$ is decomposed into two parts:

• $G(D)z=q$ (a linear differential equation of $z$, given $q$),
• $H(D)y=z$ (a linear differential equation of $y$, given $z$),

and we can process one after the other. Thus, by factorizing the polynomial $F(D)$, we only need to consider the case where $F(t)=t-\alpha$ ($\alpha \in \mathbb{C}$). That is, $y^{\prime }-\alpha y=q.$ This ODE can be readily solved by using the method of variation of parameters. The general solution of this ODE is obtained as

$$y=e^{\alpha x}\{\int e^{-\alpha x}q(x)dx+C\}$$

where $C$ is a constant.

See also: First-order linear differential equations

Example. Let us solve $y^{″}+y^{\prime }-6y=\cos x$.

Note that $F(t)=t^2+t-6=(t-2)(t+3)$ so the given ODE is $(D-2)(D+3)y=\cos x$. Solving the homogeneous ODE $(D-2)(D+3)y=0$, we have

  $y=A{e}^{2x}+B{e}^{-3x}$ where $A,B$ are constants. Let $z=(D+3)y$, then we have $(D-2)z=\cos x$ and its special solution is given by

$$z=e^{2x}\int e^{-2x}\cos xdx=-\frac{2}{5}\cos x+\frac{1}{5}\sin x.$$

Similarly, solving $(D+3)y=-\frac{2}{5}\cos x+\frac{1}{5}\sin x$, we have a special solution

$$y=e^{-3x}\int e^{3x}(-\frac{2}{5}\cos x+\frac{1}{5}\sin x)dx=-\frac{7}{50}\cos x+\frac{1}{50}\sin x.$$

Thus, the general solution is

$$y=-\frac{7}{50}\cos x+\frac{1}{50}\sin x+A{e}^{2x}+B{e}^{-3x}$$

where $A,B$ are constants. □

Let's consider the more general case where $F(t)=t^2+at+b=(t-\alpha )(t-\beta )$ with $\alpha \ne \beta$. In particular, when $a^2-4b<0$, we have $\beta =\overline{\alpha }$. We want to solve the following second-order ODE:

$$(D-\alpha )(D-\beta )y=q.$$

Let $z=(D-\beta )y$, and we have

$$\begin{array}{}\text{(eq:odea)}& (D-\alpha )z=q.\end{array}$$

We first solve the homogeneous case $(D-\alpha )z=0$ which yields

$$z=A{e}^{\alpha x}.$$

Now, regarding $A$ as a function of $x$, substitute it into (eq:odea), we have

$$A^{\prime }(x)=e^{-\alpha x}q(x).$$

Thus, we have the special solution

$$\begin{array}{rcl}z& =& e^{\alpha x}A(x)\\ \text{(eq:odez)}& & =& e^{\alpha x}\int e^{-\alpha x}q(x)dx.\end{array}$$

Next, we solve

$$\begin{array}{}\text{(eq:odeb)}& (D-\beta )y=z.\end{array}$$

Solving the homogeneous case $(D-\beta )y=0$, we have

$$y=B{e}^{\beta x}.$$

Regarding $B$ as a function of $x$ and substituting it into (eq:odeb), we have

$$B^{\prime }(x)=e^{-\beta x}z(x).$$

Integrating this using (eq:odea), 

$$\begin{array}{rcl}B(x)& =& \int e^{-\beta x}{e}^{\alpha x}A(x)dx\\ & =& \frac{e^{(\alpha -\beta )x}}{\alpha -\beta }A(x)-\int \frac{e^{(\alpha -\beta )x}}{\alpha -\beta }{A}^{\prime }(x)dx\text{(integration by parts)}\\ & =& \frac{e^{(\alpha -\beta )x}}{\alpha -\beta }A(x)+\int \frac{e^{(\alpha -\beta )x}}{\beta -\alpha }{e}^{-\alpha x}q(x)dx\\ & =& \frac{e^{(\alpha -\beta )x}}{\alpha -\beta }\int e^{-\alpha x}q(x)dx+\frac{1}{\beta -\alpha }\int e^{-\beta x}q(x)dx.\end{array}$$

Thus, we have the special solution

$$\begin{array}{rcl}y& =& e^{\beta x}B(x)\\ & =& \frac{e^{\alpha x}}{\alpha -\beta }\int e^{-\alpha x}q(x)dx+\frac{e^{\beta x}}{\beta -\alpha }\int e^{-\beta x}q(x)dx.\end{array}$$

Note the symmetry between the two terms on the right-hand side. This means that swapping $\alpha$ and $\beta$ does not change the result. This is expected as the differential operators are commutative: $(D-\alpha )(D-\beta )=(D-\beta )(D-\alpha )$.

Example. Let us solve

$$y^{″}+y=\frac{1}{\cos x}.$$

The corresponding homogeneous equation is $y^{″}+y=0$. Noting $D^2+1=(D-i)(D+i)$, its solution is

$$\begin{array}{}\text{(eg:hom)}& y=A{e}^{ix}+B{e}^{-ix}\end{array}$$

where $A,B\in \mathbb{C}$ are constants. Now, we use the method of variation of parameters by regarding $A$ and $B$ as functions of $x$. According to the above result, we have

$$\begin{array}{rcl}A(x)& =& \int \frac{e^{-ix}}{2i\cos x}dx\\ & =& \frac{1}{2i}\int (1-i\frac{\sin x}{\cos x})dx\\ & =& \frac{1}{2i}(x+i\log |\cos x|)+C_1\end{array}$$

where $C_1$ is a constant. Similarly,

$$B(x)=-\frac{1}{2i}(x-i\log |\cos x|)+C_2$$

where $C_2$ is a constant. Thus, the general solution is

$$y=[\frac{1}{2i}(x+i\log |\cos x|)+C_1]e^{ix}+[-\frac{1}{2i}(x-i\log |\cos x|)+C_2]e^{-ix}.$$

Using Euler's formula ($e^{\pm ix}=\cos x\pm i\sin x$), this can be further "simplified" (exercise!) as

$$y=(\log |\cos x|+C)\cos x+(x+D)\sin x$$

where $C$ and $D$ are constants. Note that this is a real-valued function if $C$ and $D$ are real. □

Next, let's consider $(D-\alpha {)}^{2}y=q.$ Of course, $(t-\alpha {)}^2=(t-\alpha )(t-\alpha )$ so that we can apply the technique above. Let $y_1=(D-\alpha )y$, and we solve $(D-\alpha )y_1=q$. As we have seen above, 

$$y_1=e^{\alpha x}\int e^{-\alpha x}q(x)dx.$$

Next, we solve $(D-\alpha )y_2=y_1$. This gives

$$y_2=e^{\alpha x}\int e^{-\alpha x}{y}_1(x)dx.$$

We can continue the same process to solve $(D-\alpha {)}^{m}y=q$ for any $m=3,4,\cdots$. Let the solution of $(D-\alpha {)}^{m}y=q$ be $y_m$ for $m=0,1,2,\cdots$. We can see that

$$y_m=e^{\alpha x}\int e^{-\alpha x}{y}_{m-1}(x)dx$$

and

$$y_0(x)=q(x).$$

Example. Solve $y^{″}-4y^{\prime }+4y=\cos x$. It is an exercise to show that the general solution is

$$y=\frac{3}{25}\cos x-\frac{4}{25}\sin x+A{e}^{2x}+Bx{e}^{2x}$$

where $A$ and $B$ are constants. □

See also: Homogeneous linear differential equations with constant coefficients