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# 2.3: Linear Second Order Nonhomogeneous Linear Equations

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We'll now consider the nonhomogeneous linear second order equation

\label{eq:2.3.1}
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 $$(2.1.1)$$, gives sufficient conditions for existence and uniqueness of solutions of initial value problems for \eqref{eq:2.3.1}. We omit the proof, which is beyond the scope of this book.

### Theorem $$\PageIndex{1}$$

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

\begin{eqnarray*}
\end{eqnarray*}

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

Proof

Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page.

To find the general solution of \eqref{eq:2.3.1} on an interval $$(a,b)$$ where $$p$$, $$q$$, and $$f$$ are continuous, it's necessary to find the general solution of the associated homogeneous equation

\label{eq:2.3.2}
y''+p(x)y'+q(x)y=0

on $$(a,b)$$. We call \eqref{eq:2.3.2} the $$\textcolor{blue}{\mbox{complementary equation}}$$ for \eqref{eq:2.3.1}.

The next theorem shows how to find the general solution of \eqref{eq:2.3.1} if we know one solution $$y_p$$ of \eqref{eq:2.3.1} and a fundamental set of solutions of \eqref{eq:2.3.2}. We call $$y_p$$ a $$\textcolor{blue}{\mbox{particular solution}}$$ of \eqref{eq:2.3.1}; it can be any solution that we can find, one way or another.

### Theorem $$\PageIndex{2}$$

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

\label{eq:2.3.3}
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:2.3.4}
y''+p(x)y'+q(x)y=0

on $$(a,b)$$. Then $$y$$ is a solution of \eqref{eq:2.3.3} on $$(a,b)$$ if and only if

\label{eq:2.3.5}
y=y_p+c_1y_1+c_2y_2,

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

Proof

We first show that $$y$$ in \eqref{eq:2.3.5} is a solution of \eqref{eq:2.3.3} for any choice of the constants $$c_1$$ and $$c_2$$. Differentiating \eqref{eq:2.3.5} twice yields

\begin{eqnarray*}
\end{eqnarray*}

so

\begin{eqnarray*}
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{eqnarray*}

since $$y_p$$ satisfies \eqref{eq:2.3.3} and $$y_1$$ and $$y_2$$ satisfy \eqref{eq:2.3.4}.

Now we'll show that every solution of \eqref{eq:2.3.3} has the form \eqref{eq:2.3.5} for some choice of the constants $$c_1$$ and $$c_2$$. Suppose $$y$$ is a solution of \eqref{eq:2.3.3}. We'll show that $$y-y_p$$ is a solution of \eqref{eq:2.3.4}, and therefore of the form $$y-y_p=c_1y_1+c_2y_2$$, which implies \eqref{eq:2.3.5}. To see this, we compute

\begin{eqnarray*}
(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{eqnarray*}

since $$y$$ and $$y_p$$ both satisfy \eqref{eq:2.3.3}.

We say that \eqref{eq:2.3.5} is the $$\textcolor{blue}{\mbox{general solution of \(\eqref{eq:2.3.3}$$ on $$(a,b)$$.}} \)

If $$P_0$$, $$P_1$$, and $$F$$ are continuous and $$P_0$$ has no zeros on $$(a,b)$$, then Theorem $$(2.3.2)$$ implies that the general solution of

\label{eq:2.3.6}
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 \eqref{eq:2.3.6} on $$(a,b)$$ and $$\{y_1,y_2\}$$ is a fundamental set of solutions of

\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=0
\end{eqnarray*}

on $$(a,b)$$. To see this, we rewrite \eqref{eq:2.3.6} as

\begin{eqnarray*}
y''+{P_1(x)\over P_0(x)}y'+{P_2(x)\over P_0(x)}y={F(x)\over P_0(x)}
\end{eqnarray*}

and apply Theorem $$(2.3.2)$$ 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 won't 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 \eqref{eq:2.3.3}, 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 \eqref{eq:2.3.6}. 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 $$P_0$$ $$\textcolor{blue}{\mbox{does}}$$ have a zero in $$(a,b)$$, then \eqref{eq:2.3.6} may fail to have a general solution on $$(a,b)$$ in the sense just defined. Exercises $$(2.1E.42)$$, $$(2.1E.43)$$, and $$(2.1E.44)$$ illustrate this point for a homogeneous equation.

In this section we limit ourselves to applications of Theorem $$(2.3.2)$$ where we can guess at the form of the particular solution.

### Example $$\PageIndex{1}$$

(a) Find the general solution of

\label{eq:2.3.7}
y''+y=1.

(b) Solve the initial value problem

\label{eq:2.3.8}

(a) We can apply Theorem $$(2.3.2)$$ with $$(a,b)= (-\infty,\infty)$$, since the functions $$p\equiv0$$, $$q\equiv1$$, and $$f\equiv1$$ in \eqref{eq:2.3.7} are continuous on $$(-\infty,\infty)$$. By inspection we see that $$y_p\equiv1$$ is a particular solution of \eqref{eq:2.3.7}. 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 \eqref{eq:2.3.7} is

\label{eq:2.3.9}
y=1+c_1\cos x+c_2\sin x.

(b) Imposing the initial condition $$y(0)=2$$ in \eqref{eq:2.3.9} yields $$2=1+c_1$$, so $$c_1=1$$. Differentiating \eqref{eq:2.3.9} yields

\begin{eqnarray*}
y'=-c_1\sin x+c_2\cos x.
\end{eqnarray*}

Imposing the initial condition $$y'(0)=7$$ here yields $$c_2=7$$, so the solution of \eqref{eq:2.3.8} is

\begin{eqnarray*}
y=1+\cos x+7\sin x.
\end{eqnarray*}

Figure $$2.3.1$$ is a graph of this function.

### Figure: $$2.3.1$$

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

### Example $$\PageIndex{2}$$

(a) Find the general solution of

\label{eq:2.3.10}
y''-2y'+y=-3-x+x^2.

(b) Solve the initial value problem

\label{eq:2.3.11}

(a) The characteristic polynomial of the complementary equation

\begin{eqnarray*}
y''-2y'+y=0
\end{eqnarray*}

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 \eqref{eq:2.3.10}, we note that substituting a second degree polynomial $$y_p=A+Bx+Cx^2$$ into the left side of \eqref{eq:2.3.10} 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 \eqref{eq:2.3.10} have the same coefficients; thus, if

\begin{eqnarray*}
\end{eqnarray*}

so

\begin{eqnarray*}
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{eqnarray*}

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

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

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 \eqref{eq:2.3.10} and Theorem $$(2.3.2)$$ implies that

\label{eq:2.3.12}
y=1+3x+x^2+e^x(c_1+c_2x)

is the general solution of \eqref{eq:2.3.10}.

(b) Imposing the initial condition $$y(0)=-2$$ in \eqref{eq:2.3.12} yields $$-2=1+c_1$$, so $$c_1=-3$$. Differentiating \eqref{eq:2.3.12} yields

\begin{eqnarray*}
y'=3+2x+e^x(c_1+c_2x)+c_2e^x,
\end{eqnarray*}

and imposing the initial condition $$y'(0)=1$$ here yields $$1=3+c_1+c_2$$, so $$c_2=1$$. Therefore the solution of \eqref{eq:2.3.11} is

\begin{eqnarray*}
y=1+3x+x^2-e^x(3-x).
\end{eqnarray*}

Figure $$2.3.2$$ is a graph of this solution.

### Figure: $$2.3.2$$

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

### Example $$\PageIndex{3}$$

Find the general solution of

\label{eq:2.3.13}
x^2y''+xy'-4y=2x^4

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

In Example $$(2.3.1)$$, we verified that $$y_1=x^2$$ and $$y_2=1/x^2$$ form a fundamental set of solutions of the complementary equation

\begin{eqnarray*}
x^2y''+xy'-4y=0
\end{eqnarray*}

on $$(-\infty,0)$$ and $$(0,\infty)$$. To find a particular solution of \eqref{eq:2.3.13}, we note that if $$y_p=Ax^4$$, where $$A$$ is a constant then both sides of \eqref{eq:2.3.13} 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

\begin{eqnarray*}
x^2y_p''+xy_p'-4y_p=x^2(12Ax^2)+x(4Ax^3)-4Ax^4=12Ax^4=2x^4
\end{eqnarray*}

if $$A=1/6$$; therefore, $$y_p=x^4/6$$ is a particular solution of \eqref{eq:2.3.13} on $$(-\infty,\infty)$$. Theorem $$(2.3.2)$$ implies that the general solution of \eqref{eq:2.3.13} on $$(-\infty,0)$$ and $$(0,\infty)$$ is

\begin{eqnarray*}
y={x^4\over6}+c_1x^2+{c_2\over x^2}.
\end{eqnarray*}

## The Principle of Superposition

The next theorem enables us to break a nonhomogeneous 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 $$\PageIndex{3}$$

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

\begin{eqnarray*}
y''+p(x)y'+q(x)y=f_1(x)
\end{eqnarray*}

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

\begin{eqnarray*}
y''+p(x)y'+q(x)y=f_2(x)
\end{eqnarray*}

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

\begin{eqnarray*}
y_p=y_{p_1}+y_{p_2}
\end{eqnarray*}

is a particular solution of

\begin{eqnarray*}
y''+p(x)y'+q(x)y=f_1(x)+f_2(x)
\end{eqnarray*}

on $$(a,b)$$.

Proof

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

\begin{eqnarray*}
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{eqnarray*}

It's easy to generalize Theorem $$(2.3.3)$$ to the equation

\label{eq:2.3.14}
y''+p(x)y'+q(x)y=f(x)

where

\begin{eqnarray*}
f=f_1+f_2+\cdots+f_k;
\end{eqnarray*}

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

\begin{eqnarray*}
y''+p(x)y'+q(x)y=f_i(x)
\end{eqnarray*}

on $$(a,b)$$ for $$i=1$$, $$2$$, $$\dots$$, $$k$$, then $$y_{p_1}+y_{p_2}+\cdots+y_{p_k}$$ is a particular solution of \eqref{eq:2.3.14} on $$(a,b)$$. Moreover, by a proof similar to the proof of Theorem $$(2.3.3)$$ we can formulate the principle of superposition in terms of a linear equation written in the form

\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=F(x)
\end{eqnarray*}

(Exercise $$(2.3E.39)$$); that is, if $$y_{p_1}$$ is a particular solution of

\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)
\end{eqnarray*}

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

\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=F_2(x)
\end{eqnarray*}

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

\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)+F_2(x)
\end{eqnarray*}

on $$(a,b)$$.

### Example $$\PageIndex{4}$$

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

\label{eq:2.3.15}
x^2y''+4xy'+2y=2x^4

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

\label{eq:2.3.16}
x^2y''+4xy'+2y=4x^2

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

\label{eq:2.3.17}
x^2y''+4xy'+2y=2x^4+4x^2

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

The right side $$F(x)=2x^4+4x^2$$ in \eqref{eq:2.3.17} is the sum of the right sides

\begin{eqnarray*}
F_1(x)=2x^4 \quad \mbox{ and } \quad F_2(x)=4x^2.
\end{eqnarray*}

in \eqref{eq:2.3.15} and \eqref{eq:2.3.16}. Therefore the principle of superposition implies that

\begin{eqnarray*}
y_p=y_{p_1}+y_{p_2}={x^4\over15}+{x^2\over3}
\end{eqnarray*}

is a particular solution of \eqref{eq:2.3.17}.