
# 5.3: Nonhomgeneous Linear Equations


We’ll now consider the nonhomogeneous linear second order equation

$\label{eq:5.3.1} y''+p(x)y'+q(x)y=f(x),$

where the forcing function $$f$$ isn’t identically zero. The next theorem, an extension of Theorem 5.1.1, gives sufficient conditions for existence and uniqueness of solutions of initial value problems for Equation \ref{eq:5.3.1}. We omit the proof, which is beyond the scope of this book.

##### Theorem 5.3.1 : Uniqueness

Suppose $$p,$$ $$,q$$ and $$f$$ are continuous on an open interval $$(a,b),$$ let $$x_0$$ be any point in $$(a,b),$$ and let $$k_0$$ and $$k_1$$ be arbitrary real numbers$$.$$ Then the initial value problem

$y''+p(x)y'+q(x)y=f(x), \quad y(x_0)=k_0,\quad y'(x_0)=k_1\nonumber$

has a unique solution on $$(a,b).$$

To find the general solution of Equation \ref{eq:5.3.1} on an interval $$(a,b)$$ where $$p$$, $$q$$, and $$f$$ are continuous, it is necessary to find the general solution of the associated homogeneous equation

$\label{eq:5.3.2} y''+p(x)y'+q(x)y=0$

on $$(a,b)$$. We call Equation \ref{eq:5.3.2} the complementary equation for Equation \ref{eq:5.3.1}.

The next theorem shows how to find the general solution of Equation \ref{eq:5.3.1} if we know one solution $$y_p$$ of Equation \ref{eq:5.3.1} and a fundamental set of solutions of Equation \ref{eq:5.3.2}. We call $$y_p$$ a particular solution of Equation \ref{eq:5.3.1} ; it can be any solution that we can find, one way or another.

##### Theorem 5.3.2

Suppose $$p,$$ $$q,$$ and $$f$$ are continuous on $$(a,b).$$ Let $$y_p$$ be a particular solution of

$\label{eq:5.3.3} y''+p(x)y'+q(x)y=f(x)$

on $$(a,b)$$, and let $$\{y_1,y_2\}$$ be a fundamental set of solutions of the complementary equation

$\label{eq:5.3.4} y''+p(x)y'+q(x)y=0$

on $$(a,b)$$. Then $$y$$ is a solution of $$\eqref{eq:5.3.3}$$ on $$(a,b)$$ if and only if

$\label{eq:5.3.5} y=y_p+c_1y_1+c_2y_2,$

where $$c_1$$ and $$c_2$$ are constants.

Proof

We first show that $$y$$ in Equation \ref{eq:5.3.5} is a solution of Equation \ref{eq:5.3.3} for any choice of the constants $$c_1$$ and $$c_2$$. Differentiating Equation \ref{eq:5.3.5} twice yields

$y'=y_p'+c_1y_1'+c_2y_2' \quad \text{and} \quad y''=y_p''+ c_1y_1''+c_2y_2'', \nonumber$

so

\begin{align*} y''+p(x)y'+q(x)y&=(y_p''+c_1y_1''+c_2y_2'') +p(x)(y_p'+c_1y_1'+c_2y_2') +q(x)(y_p+c_1y_1+c_2y_2)\\ &=(y_p''+p(x)y_p'+q(x)y_p)+c_1(y_1''+p(x)y_1'+q(x)y_1) +c_2(y_2''+p(x)y_2'+q(x)y_2)\\ &= f+c_1\cdot0+c_2\cdot0=f,\end{align*}

since $$y_p$$ satisfies Equation \ref{eq:5.3.3} and $$y_1$$ and $$y_2$$ satisfy Equation \ref{eq:5.3.4}.

Now we’ll show that every solution of Equation \ref{eq:5.3.3} has the form Equation \ref{eq:5.3.5} for some choice of the constants $$c_1$$ and $$c_2$$. Suppose $$y$$ is a solution of Equation \ref{eq:5.3.3}. We’ll show that $$y-y_p$$ is a solution of Equation \ref{eq:5.3.4}, and therefore of the form $$y-y_p=c_1y_1+c_2y_2$$, which implies Equation \ref{eq:5.3.5}. To see this, we compute

\begin{align*} (y-y_p)''+p(x)(y-y_p)'+q(x)(y-y_p)&=(y''-y_p'')+p(x)(y'-y_p') +q(x)(y-y_p)\\ &=(y''+p(x)y'+q(x)y) -(y_p''+p(x)y_p'+q(x)y_p)\\ &=f(x)-f(x)=0,\end{align*}

since $$y$$ and $$y_p$$ both satisfy Equation \ref{eq:5.3.3}.

We say that Equation \ref{eq:5.3.5} is the general solution of $$\eqref{eq:5.3.3}$$ on $$(a,b)$$.

If $$P_0$$, $$P_1$$, and $$F$$ are continuous and $$P_0$$ has no zeros on $$(a,b)$$, then Theorem 5.3.2 implies that the general solution of

$\label{eq:5.3.6} P_0(x)y''+P_1(x)y'+P_2(x)y=F(x)$

on $$(a,b)$$ is $$y=y_p+c_1y_1+c_2y_2$$, where $$y_p$$ is a particular solution of Equation \ref{eq:5.3.6} on $$(a,b)$$ and $$\{y_1,y_2\}$$ is a fundamental set of solutions of

$P_0(x)y''+P_1(x)y'+P_2(x)y=0\nonumber$

on $$(a,b)$$. To see this, we rewrite Equation \ref{eq:5.3.6} as

$y''+{P_1(x)\over P_0(x)}y'+{P_2(x)\over P_0(x)}y={F(x)\over P_0(x)}\nonumber$

and apply Theorem 5.3.2 with $$p=P_1/P_0$$, $$q=P_2/P_0$$, and $$f=F/P_0$$.

To avoid awkward wording in examples and exercises, we will not specify the interval $$(a,b)$$ when we ask for the general solution of a specific linear second order equation, or for a fundamental set of solutions of a homogeneous linear second order equation. Let’s agree that this always means that we want the general solution (or a fundamental set of solutions, as the case may be) on every open interval on which $$p$$, $$q$$, and $$f$$ are continuous if the equation is of the form Equation \ref{eq:5.3.3}, or on which $$P_0$$, $$P_1$$, $$P_2$$, and $$F$$ are continuous and $$P_0$$ has no zeros, if the equation is of the form Equation \ref{eq:5.3.6}. We leave it to you to identify these intervals in specific examples and exercises.

For completeness, we point out that if $$P_0$$, $$P_1$$, $$P_2$$, and $$F$$ are all continuous on an open interval $$(a,b)$$, but $$P_0$$ does have a zero in $$(a,b)$$, then Equation \ref{eq:5.3.6} may fail to have a general solution on $$(a,b)$$ in the sense just defined. Exercises 5.1.42-5.1.44 illustrate this point for a homogeneous equation.

In this section we to limit ourselves to applications of Theorem 5.3.2 where we can guess at the form of the particular solution.

##### Example 5.3.1
1. Find the general solution of $\label{eq:5.3.7} y''+y=1.$
2. Solve the initial value problem $\label{eq:5.3.8} y''+y=1, \quad y(0)=2,\quad y'(0)=7.$

Solution

a. We can apply Theorem 5.3.2 with $$(a,b)= (-\infty,\infty)$$, since the functions $$p\equiv0$$, $$q\equiv1$$, and $$f\equiv1$$ in Equation \ref{eq:5.3.7} are continuous on $$(-\infty,\infty)$$. By inspection we see that $$y_p\equiv1$$ is a particular solution of Equation \ref{eq:5.3.7}. Since $$y_1=\cos x$$ and $$y_2=\sin x$$ form a fundamental set of solutions of the complementary equation $$y''+y=0$$, the general solution of Equation \ref{eq:5.3.7} is

$\label{eq:5.3.9} y=1+c_1\cos x + c_2\sin x.$

b. Imposing the initial condition $$y(0)=2$$ in Equation \ref{eq:5.3.9} yields $$2=1+c_1$$, so $$c_1=1$$. Differentiating Equation \ref{eq:5.3.9} yields

$y'=-c_1\sin x+c_2\cos x.\nonumber$

Imposing the initial condition $$y'(0)=7$$ here yields $$c_2=7$$, so the solution of Equation \ref{eq:5.3.8} is

$y=1+\cos x+7\sin x.\nonumber$

Figure 5.3.1 is a graph of this function.

##### Example 5.3.2
1. Find the general solution of $\label{eq:5.3.10} y''-2y'+y=-3-x+x^2.$
2. Solve the initial value problem $\label{eq:5.3.11} y''-2y'+y=-3-x+x^2, \quad y(0)=-2,\quad y'(0)=1.$

Solution

a. The characteristic polynomial of the complementary equation

$y''-2y'+y=0 \nonumber$

is $$r^2-2r+1=(r-1)^2$$, so $$y_1=e^x$$ and $$y_2=xe^x$$ form a fundamental set of solutions of the complementary equation. To guess a form for a particular solution of Equation \ref{eq:5.3.10}, we note that substituting a second degree polynomial $$y_p=A+Bx+Cx^2$$ into the left side of Equation \ref{eq:5.3.10} will produce another second degree polynomial with coefficients that depend upon $$A$$, $$B$$, and $$C$$. The trick is to choose $$A$$, $$B$$, and $$C$$ so the polynomials on the two sides of Equation \ref{eq:5.3.10} have the same coefficients; thus, if

$y_p=A+Bx+Cx^2 \quad \text{then} \quad y_p'=B+2Cx \quad \text{and} \quad y_p''=2C, \nonumber$

so

\begin{aligned} y_p''-2y_p'+y_p&=2C-2(B+2Cx)+(A+Bx+Cx^2)\\ &=(2C-2B+A)+(-4C+B)x+Cx^2=-3-x+x^2.\end{aligned}\nonumber

Equating coefficients of like powers of $$x$$ on the two sides of the last equality yields

\begin{align*} C&= \phantom{-}1\phantom{.}\\ B-4C&=-1\phantom{.}\\ A-2B+2C&= -3,\end{align*}

so $$C=1$$, $$B=-1+4C=3$$, and $$A=-3-2C+2B=1$$. Therefore $$y_p=1+3x+x^2$$ is a particular solution of Equation \ref{eq:5.3.10} and Theorem 5.3.2 implies that

$\label{eq:5.3.12} y=1+3x+x^2+e^x(c_1+c_2x)$

is the general solution of Equation \ref{eq:5.3.10}.

b. Imposing the initial condition $$y(0)=-2$$ in Equation \ref{eq:5.3.12} yields $$-2=1+c_1$$, so $$c_1=-3$$. Differentiating Equation \ref{eq:5.3.12} yields

$y'=3+2x+e^x(c_1+c_2x)+c_2e^x,\nonumber$

and imposing the initial condition $$y'(0)=1$$ here yields $$1=3+c_1+c_2$$, so $$c_2=1$$. Therefore the solution of Equation \ref{eq:5.3.11} is

$y=1+3x+x^2-e^x(3-x). \nonumber$

Figure 5.3.2 is a graph of this solution.

##### Example 5.3.3

Find the general solution of

$\label{eq:5.3.13} x^2y''+xy'-4y=2x^4$

on $$(-\infty,0)$$ and $$(0,\infty)$$.

Solution

In Example 5.1.3, we verified that $$y_1=x^2$$ and $$y_2=1/x^2$$ form a fundamental set of solutions of the complementary equation

$x^2y''+xy'-4y=0 \nonumber$

on $$(-\infty,0)$$ and $$(0,\infty)$$. To find a particular solution of Equation \ref{eq:5.3.13}, we note that if $$y_p=Ax^4$$, where $$A$$ is a constant then both sides of Equation \ref{eq:5.3.13} will be constant multiples of $$x^4$$ and we may be able to choose $$A$$ so the two sides are equal. This is true in this example, since if $$y_p=Ax^4$$ then

$x^2y_p''+xy_p'-4y_p=x^2(12Ax^2)+x(4Ax^3)-4Ax^4=12Ax^4=2x^4 \nonumber$

if $$A=1/6$$; therefore, $$y_p=x^4/6$$ is a particular solution of Equation \ref{eq:5.3.13} on $$(-\infty,\infty)$$. Theorem 5.3.2 implies that the general solution of Equation \ref{eq:5.3.13} on $$(-\infty,0)$$ and $$(0,\infty)$$ is

$y={x^4\over6}+c_1x^2+{c_2\over x^2}. \nonumber$

## The Principle of Superposition

The next theorem enables us to break a nonhomogeous equation into simpler parts, find a particular solution for each part, and then combine their solutions to obtain a particular solution of the original problem.

##### Theorem 5.3.3 the principle of superposition

Suppose $$y_{p_1}$$ is a particular solution of

$y''+p(x)y'+q(x)y=f_1(x) \nonumber$

on $$(a,b)$$ and $$y_{p_2}$$ is a particular solution of

$y''+p(x)y'+q(x)y=f_2(x) \nonumber$

on $$(a,b)$$. Then

$y_p=y_{p_1}+y_{p_2} \nonumber$

is a particular solution of

$y''+p(x)y'+q(x)y=f_1(x)+f_2(x) \nonumber$

on $$(a,b)$$.

Proof

If $$y_p=y_{p_1}+y_{p_2}$$ then

\begin{align*} y_p''+p(x)y_p'+q(x)y_p&=(y_{p_1}+y_{p_2})''+p(x)(y_{p_1}+y_{p_2})' +q(x)(y_{p_1}+y_{p_2})\\ &=\left(y_{p_1}''+p(x)y_{p_1}'+q(x)y_{p_1}\right) +\left(y_{p_2}''+p(x)y_{p_2}'+q(x)y_{p_2}\right)\\ &=f_1(x)+f_2(x). \end{align*}

It’s easy to generalize Theorem 5.3.3 to the equation

$\label{eq:5.3.14} y''+p(x)y'+q(x)y=f(x)$

where

$f=f_1+f_2+\cdots+f_k; \nonumber$

thus, if $$y_{p_i}$$ is a particular solution of

$y''+p(x)y'+q(x)y=f_i(x) \nonumber$

on $$(a,b)$$ for $$i=1$$, $$2$$, …, $$k$$, then $$y_{p_1}+y_{p_2}+\cdots+y_{p_k}$$ is a particular solution of Equation \ref{eq:5.3.14} on $$(a,b)$$. Moreover, by a proof similar to the proof of Theorem 5.3.3 we can formulate the principle of superposition in terms of a linear equation written in the form

$P_0(x)y''+P_1(x)y'+P_2(x)y=F(x) \nonumber$

(Exercise 5.3.39); that is, if $$y_{p_1}$$ is a particular solution of

$P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x) \nonumber$

on $$(a,b)$$ and $$y_{p_2}$$ is a particular solution of

$P_0(x)y''+P_1(x)y'+P_2(x)y=F_2(x) \nonumber$

on $$(a,b)$$, then $$y_{p_1}+y_{p_2}$$ is a solution of

$P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)+F_2(x) \nonumber$

on $$(a,b)$$.

##### Example 5.3.4

The function $$y_{p_1}=x^4/15$$ is a particular solution of

$\label{eq:5.3.15} x^2y''+4xy'+2y=2x^4$

on $$(-\infty,\infty)$$ and $$y_{p_2}=x^2/3$$ is a particular solution of

$\label{eq:5.3.16} x^2y''+4xy'+2y=4x^2$

on $$(-\infty,\infty)$$. Use the principle of superposition to find a particular solution of

$\label{eq:5.3.17} x^2y''+4xy'+2y=2x^4+4x^2$

on $$(-\infty,\infty)$$.

Solution

The right side $$F(x)=2x^4+4x^2$$ in Equation \ref{eq:5.3.17} is the sum of the right sides

$F_1(x)=2x^4\quad \text{and} \quad F_2(x)=4x^2. \nonumber$

in Equation \ref{eq:5.3.15} and Equation \ref{eq:5.3.16}. Therefore the principle of superposition implies that

$y_p=y_{p_1}+y_{p_2}={x^4\over15}+{x^2\over3} \nonumber$

is a particular solution of Equation \ref{eq:5.3.17}.

5.3: Nonhomgeneous Linear Equations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.