5.5: The Method of Undetermined Coefficients II

In this section we consider the constant coefficient equation

$$\label{eq:5.5.1} ay''+by'+cy=e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)$$

where $\lambda$ and $\omega$ are real numbers, $\omega\ne0$, and $P$ and $Q$ are polynomials. We want to find a particular solution of Equation $\ref{eq:5.5.1}$. As in Section 5.4, the procedure that we will use is called the method of undetermined coefficients.

Forcing Functions Without Exponential Factors

We begin with the case where $\lambda=0$ in Equation $\ref{eq:5.5.1}$ ; thus, we we want to find a particular solution of

$$\label{eq:5.5.2} ay''+by'+cy=P(x)\cos\omega x+Q(x)\sin\omega x,$$

where $P$ and $Q$ are polynomials.

Differentiating $x^r\cos\omega x$ and $x^r\sin\omega x$ yields

$${d\over dx}x^r\cos\omega x=-\omega x^r\sin\omega x+ rx^{r-1}\cos\omega x \nonumber$$

and

$${d\over dx}x^r\sin\omega x=\phantom{-}\omega x^r\cos\omega x+ rx^{r-1}\sin\omega x. \nonumber$$

This implies that if

$$y_p=A(x)\cos\omega x+B(x)\sin\omega x\nonumber$$

where $A$ and $B$ are polynomials, then

$$ay_p''+by_p'+cy_p=F(x)\cos\omega x+G(x)\sin\omega x,\nonumber$$

where $F$ and $G$ are polynomials with coefficients that can be expressed in terms of the coefficients of $A$ and $B$. This suggests that we try to choose $A$ and $B$ so that $F=P$ and $G=Q$, respectively. Then $y_p$ will be a particular solution of Equation $\ref{eq:5.5.2}$. The next theorem tells us how to choose the proper form for $y_p$. For the proof see Exercise 5.5.37.

Theorem 5.5.1

Suppose $\omega$ is a positive number and $P$ and $Q$ are polynomials. Let $k$ be the larger of the degrees of $P$ and $Q.$ Then the equation

$$ay''+by'+cy=P(x)\cos \omega x+Q(x)\sin \omega x \nonumber$$

has a particular solution

$$\label{eq:5.5.3} y_p=A(x)\cos\omega x+B(x)\sin\omega x,$$

where

$$A(x)=A_0+A_1x+\cdots+A_kx^k \quad \text{and} \quad B(x)=B_0+B_1x+\cdots+B_kx^k, \nonumber$$

provided that $\cos\omega x$ and $\sin\omega x$ are not solutions of the complementary equation. The solutions of

$$a(y''+\omega^2y)=P(x)\cos \omega x+Q(x)\sin \omega x \nonumber$$

for which $\cos\omega x$ and $\sin\omega x$ are solutions of the complementary equation are of the form of Equation $\ref{eq:5.5.3}$, where

$$A(x)=A_0x+A_1x^2+\cdots+A_kx^{k+1} \quad \text{and} \quad B(x)=B_0x+B_1x^2+\cdots+B_kx^{k+1}. \nonumber$$

For an analog of this theorem that’s applicable to Equation $\ref{eq:5.5.1}$, see Exercise 5.5.38.

Example 5.5.1

Find a particular solution of

$$\label{eq:5.5.4} y''-2y'+y=5\cos2x+10\sin2x.$$

Solution

In Equation $\ref{eq:5.5.4}$ the coefficients of $\cos2x$ and $\sin2x$ are both zero degree polynomials (constants). Therefore Theorem 5.5.1 implies that Equation $\ref{eq:5.5.4}$ has a particular solution

$$y_p=A\cos2x+B\sin2x.\nonumber$$

Since

$$y_p'=-2A\sin2x+2B\cos2x\quad \text{and} \quad y_p''=-4(A\cos2x+B\sin2x),\nonumber$$

replacing $y$ by $y_p$ in Equation $\ref{eq:5.5.4}$ yields

$$\begin{aligned} y_p''-2y_p'+y_p&=-4(A\cos2x+B\sin2x)-4(-A\sin2x+B\cos2x) +(A\cos2x+B\sin2x)\\[4pt] &= (-3A-4B)\cos2x+(4A-3B)\sin2x.\end{aligned}\nonumber$$

Equating the coefficients of $\cos2x$ and $\sin2x$ here with the corresponding coefficients on the right side of Equation $\ref{eq:5.5.4}$ shows that $y_p$ is a solution of Equation $\ref{eq:5.5.4}$ if

$$\begin{aligned} -3A-4B&=\phantom{1}5\phantom{.}\\[4pt] \phantom{-}4A-3B&=10.\end{aligned}\nonumber$$

Solving these equations yields $A=1$, $B=-2$. Therefore

$$y_p=\cos2x-2\sin2x\nonumber$$

is a particular solution of Equation $\ref{eq:5.5.4}$.

Example 5.5.2

Find a particular solution of

$$\label{eq:5.5.5} y''+4y=8\cos2x+12\sin2x.$$

Solution

The procedure used in Example 5.5.1 doesn’t work here; substituting $y_p=A\cos2x+B\sin2x$ for $y$ in Equation $\ref{eq:5.5.5}$ yields

$$y_p''+4y_p=-4(A\cos2x+B\sin2x) +4(A\cos2x+B\sin2x)=0\nonumber$$

for any choice of $A$ and $B$, since $\cos2x$ and $\sin2x$ are both solutions of the complementary equation for Equation $\ref{eq:5.5.5}$. We’re dealing with the second case mentioned in Theorem 5.5.1 , and should therefore try a particular solution of the form

$$\label{eq:5.5.6} y_p=x(A\cos2x+B\sin2x).$$

Then

$$\begin{aligned} y_p'&=A\cos2x+B\sin2x+2x(-A\sin2x+B\cos2x) \quad \text{and} \\[4pt] y_p''&=-4A\sin2x+4B\cos2x-4x(A\cos2x+B\sin2x)\\[4pt] &=-4A\sin2x+4B\cos2x-4y_p \mbox{ (see \eqref{eq:5.5.6})},\end{aligned}\nonumber$$

so

$$y_p''+4y_p=-4A\sin2x+4B\cos2x.\nonumber$$

Therefore $y_p$ is a solution of Equation $\ref{eq:5.5.5}$ if

$$-4A\sin2x+4B\cos2x=8\cos2x+12\sin2x,\nonumber$$

which holds if $A=-3$ and $B=2$. Therefore

$$y_p=-x(3\cos2x-2\sin2x)\nonumber$$

is a particular solution of Equation $\ref{eq:5.5.5}$.

Example 5.5.3

Find a particular solution of

$$\label{eq:5.5.7} y''+3y'+2y=(16+20x)\cos x+10\sin x.$$

Solution

The coefficients of $\cos x$ and $\sin x$ in Equation $\ref{eq:5.5.7}$ are polynomials of degree one and zero, respectively. Therefore Theorem 5.5.1 tells us to look for a particular solution of Equation $\ref{eq:5.5.7}$ of the form

$$\label{eq:5.5.8} y_p=(A_0+A_1x)\cos x+(B_0+B_1x)\sin x.$$

Then

$$\label{eq:5.5.9} y_p'=(A_1+B_0+B_1x)\cos x+(B_1-A_0-A_1x)\sin x$$

and

$$\label{eq:5.5.10} y_p''=(2B_1-A_0-A_1x)\cos x-(2A_1+B_0+B_1x)\sin x,$$

so

$$\label{eq:5.5.11} \begin{array}{rcl} y_p''+3y_p'+2y_p&=\left[A_0+3 A_1+3 B_0+2 B_1+(A_1+3 B_1)x\right]\cos x + \left[B_0+3 B_1-3 A_0-2 A_1+(B_1-3 A_1)x\right]\sin x. \end{array}$$

Comparing the coefficients of $x\cos x$, $x\sin x$, $\cos x$, and $\sin x$ here with the corresponding coefficients in Equation $\ref{eq:5.5.7}$ shows that $y_p$ is a solution of Equation $\ref{eq:5.5.7}$ if

$$\begin{array}{rcr} \phantom{-3}A_1+3B_1&=20\phantom{.}\\[4pt] -3A_1+\phantom{3}B_1&=0\phantom{.}\\[4pt] \phantom{-3}A_0+3B_0+3A_1+2B_1&=16\phantom{.}\\[4pt] -3A_0+\phantom{3}B_0-2A_1+3B_1&=10. \end{array}\nonumber$$

Solving the first two equations yields $A_1=2$, $B_1=6$. Substituting these into the last two equations yields

$$\begin{aligned} \phantom{-3}A_0+3B_0&=16-3A_1-2B_1=-2\phantom{.}\\[4pt] -3A_0+\phantom{3}B_0&=10+2A_1-3B_1=-4. \end{aligned}\nonumber$$

Solving these equations yields $A_0=1$, $B_0=-1$. Substituting $A_0=1$, $A_1=2$, $B_0=-1$, $B_1=6$ into Equation $\ref{eq:5.5.8}$ shows that
$$y_p=(1+2x)\cos x-(1-6x)\sin x \nonumber$$

is a particular solution of Equation $\ref{eq:5.5.7}$.

A Useful Observation

In Equations $\ref{eq:5.5.9}$, $\ref{eq:5.5.10}$, and $\ref{eq:5.5.11}$ the polynomials multiplying $\sin x$ can be obtained by replacing $A_0,A_1,B_0$, and $B_1$ by $B_0$, $B_1$, $-A_0$, and $-A_1$, respectively, in the polynomials multiplying $\cos x$. An analogous result applies in general, as follows (Exercise 5.5.36).

Theorem 5.5.2

If

$$y_p=A(x)\cos\omega x+B(x)\sin\omega x, \nonumber$$

where $A(x)$ and $B(x)$ are polynomials with coefficients $A_0$ …, $A_k$ and $B_0$, …, $B_k,$ then the polynomials multiplying $\sin\omega x$ in

$$y_p',\quad y_p'',\quad ay_p''+by_p'+cy_p \quad \text{and} \quad y_p''+\omega^2 y_p \nonumber$$

can be obtained by replacing $A_0$, …$,$ $A_k$ by $B_0,$ …$,$ $B_k$ and $B_0,$ …$,$ $B_k$ by $-A_0,$ …$,$ $-A_k$ in the corresponding polynomials multiplying $\cos\omega x$.

We will not use this theorem in our examples, but we recommend that you use it to check your manipulations when you work the exercises.

Example 5.5.4

Find a particular solution of

$$\label{eq:5.5.12} y''+y=(8-4x)\cos x-(8+8x)\sin x.$$

Solution

According to Theorem 5.5.1 , we should look for a particular solution of the form

$$\label{eq:5.5.13} y_p=(A_0x+A_1x^2)\cos x+(B_0x+B_1x^2)\sin x,$$

since $\cos x$ and $\sin x$ are solutions of the complementary equation. However, let’s try

$$\label{eq:5.5.14} y_p=(A_0+A_1x)\cos x+(B_0+B_1x)\sin x$$

first, so you can see why it doesn’t work. From Equation $\ref{eq:5.5.10}$,

$$y_p''=(2B_1-A_0-A_1x)\cos x-(2A_1+B_0+B_1x)\sin x, \nonumber$$

which together with Equation $\ref{eq:5.5.14}$ implies that

$$y_p''+y_p=2B_1\cos x-2A_1\sin x. \nonumber$$

Since the right side of this equation does not contain $x\cos x$ or $x\sin x$, Equation $\ref{eq:5.5.14}$ can’t satisfy Equation $\ref{eq:5.5.12}$ no matter how we choose $A_0$, $A_1$, $B_0$, and $B_1$.

Now let $y_p$ be as in Equation $\ref{eq:5.5.13}$. Then

$$\begin{aligned} y_p'&=\left[A_0+(2A_1+B_0)x+B_1x^2\right]\cos x\\[4pt] & +\left[B_0+(2B_1-A_0)x-A_1x^2\right]\sin x \end{aligned}\nonumber$$

and

$$\begin{aligned} y_p''&= \left[2A_1+2B_0-(A_0-4B_1)x-A_1x^2\right]\cos x\\[4pt] &+ \left[2B_1-2A_0-(B_0+4A_1)x-B_1x^2\right]\sin x,\end{aligned}\nonumber$$

so

$$y_p''+y_p=(2A_1+2B_0+4B_1x)\cos x+(2B_1-2A_0-4A_1x)\sin x. \nonumber$$

Comparing the coefficients of $\cos x$ and $\sin x$ here with the corresponding coefficients in Equation $\ref{eq:5.5.12}$ shows that $y_p$ is a solution of Equation $\ref{eq:5.5.12}$ if

$$\begin{array}{rcr} \phantom{-}4B_1&=-4\phantom{.}\\[4pt] -4A_1&=-8\phantom{.}\\[4pt] \phantom{-}2B_0+2A_1&=8\phantom{.}\\[4pt] -2A_0+2B_1&=-8. \end{array}\nonumber$$

The solution of this system is $A_1=2$, $B_1=-1$, $A_0=3$, $B_0=2$. Therefore

$$y_p=x\left[(3+2x)\cos x+(2-x)\sin x\right] \nonumber$$

is a particular solution of Equation $\ref{eq:5.5.12}$.

Forcing Functions with Exponential Factors

To find a particular solution of

$$\label{eq:5.5.15} ay''+by'+cy=e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)$$

when $\lambda\ne0$, we recall from Section 5.4 that substituting $y=ue^{\lambda x}$ into Equation $\ref{eq:5.5.15}$ will produce a constant coefficient equation for $u$ with the forcing function $P(x)\cos \omega x+Q(x)\sin \omega x$. We can find a particular solution $u_p$ of this equation by the procedure that we used in Examples 5.5.1 -5.5.4 . Then $y_p=u_pe^{\lambda x}$ is a particular solution of Equation $\ref{eq:5.5.15}$.

Example 5.5.5

Find a particular solution of

$$\label{eq:5.5.16} y''-3y'+2y=e^{-2x}\left[2\cos 3x-(34-150x)\sin 3x\right].$$

Let $y=ue^{-2x}$. Then

$$\begin{aligned} y''-3y'+2y&=e^{-2x}\left[(u''-4u'+4u)-3(u'-2u)+2u\right]\\[4pt] &=e^{-2x}(u''-7u'+12u)\\[4pt] &= e^{-2x}\left[2\cos 3x-(34-150x)\sin 3x\right]\end{aligned}\nonumber$$

if

$$\label{eq:5.5.17} u''-7u'+12u=2\cos 3x-(34-150x)\sin 3x.$$

Since $\cos3x$ and $\sin3x$ aren’t solutions of the complementary equation
$$u''-7u'+12u=0,\nonumber$$

Theorem 5.5.1 tells us to look for a particular solution of Equation $\ref{eq:5.5.17}$ of the form

$$\label{eq:5.5.18} u_p=(A_0+A_1x)\cos 3x +(B_0+B_1x)\sin 3x.$$

Then

$$\begin{aligned} u_p'&=(A_1+3B_0+3B_1x)\cos 3x+(B_1-3A_0-3A_1x)\sin 3x\\[4pt] \text{and} \qquad u_p''&=(-9A_0+6B_1-9A_1x)\cos 3x-(9B_0+6A_1+9B_1x)\sin 3x,\end{aligned}\nonumber$$

so

$$\begin{aligned} u_p''-7u_p'+12u_p&=\left[3A_0-21B_0-7A_1+6B_1+(3A_1-21B_1)x\right]\cos 3x\\[4pt] &+\left[21A_0+3B_0-6A_1-7B_1+(21A_1+3B_1)x\right]\sin 3x.\end{aligned}\nonumber$$

Comparing the coefficients of $x\cos 3x$, $x\sin 3x$, $\cos 3x$, and $\sin 3x$ here with the corresponding coefficients on the right side of Equation $\ref{eq:5.5.17}$ shows that $u_p$ is a solution of Equation $\ref{eq:5.5.17}$ if

$$\label{eq:5.5.19} \begin{array}{rcr} 3A_1-21B_1&=0\phantom{.}\\[4pt] 21A_1+\phantom{2}3B_1&=150\phantom{.}\\[4pt] 3A_0-21B_0-7A_1+\phantom{2}6B_1&=\phantom{-3}2\phantom{.}\\[4pt] 21A_0+\phantom{2}3B_0-6A_1-\phantom{5}7B_1&=-34. \end{array}$$

Solving the first two equations yields $A_1=7$, $B_1=1$. Substituting these values into the last two equations of Equation $\ref{eq:5.5.19}$ yields

$$\begin{aligned} \phantom{2}3A_0-21B_0&=\phantom{-3}2+7A_1-6B_1=45\phantom{.}\\[4pt] 21A_0+\phantom{2}3B_0&=-34+6A_1+7B_1=15.\end{aligned}\nonumber$$

Solving this system yields $A_0=1$, $B_0=-2$. Substituting $A_0=1$, $A_1=7$, $B_0=-2$, and $B_1=1$ into Equation $\ref{eq:5.5.18}$ shows that

$$u_p=(1+7x)\cos 3x-(2-x)\sin 3x\nonumber$$

is a particular solution of Equation $\ref{eq:5.5.17}$. Therefore

$$y_p=e^{-2x}\left[(1+7x)\cos 3x-(2-x)\sin 3x\right]\nonumber$$

is a particular solution of Equation $\ref{eq:5.5.16}$.

Example 5.5.6

Find a particular solution of

$$\label{eq:5.5.20} y''+2y'+5y=e^{-x}\left[(6-16x)\cos2x-(8+8x)\sin2x\right].$$

Solution

Let $y=ue^{-x}$. Then

$$\begin{aligned} y''+2y'+5y&=e^{-x}\left[(u''-2u'+u)+2(u'-u)+5u\right]\\[4pt] &=e^{-x}(u''+4u)\\[4pt] &= e^{-x}\left[(6-16x)\cos2x-(8+8x)\sin2x\right]\end{aligned}\nonumber$$

if

$$\label{eq:5.5.21} u''+4u=(6-16x)\cos2x-(8+8x)\sin2x.$$

Since $\cos2x$ and $\sin2x$ are solutions of the complementary equation

$$u''+4u=0,\nonumber$$

Theorem 5.5.1 tells us to look for a particular solution of Equation $\ref{eq:5.5.21}$ of the form

$$u_p=(A_0x+A_1x^2)\cos2x+(B_0x+B_1x^2)\sin2x.\nonumber$$

Then
$$\begin{aligned} u_p'&=\left[A_0+(2A_1+2B_0)x+2B_1x^2\right]\cos2x \\[4pt] & +\left[B_0+(2B_1-2A_0)x-2A_1x^2\right]\sin2x\end{aligned}\nonumber$$

and

$$\begin{aligned} u_p''&=\left[2A_1+4B_0-(4A_0-8B_1)x-4A_1x^2\right]\cos2x\\[4pt] & +\left[2B_1-4A_0-(4B_0+8A_1)x-4B_1x^2\right]\sin2x,\end{aligned}\nonumber$$

so

$$u_p''+4u_p=(2A_1+4B_0+8B_1x)\cos2x+(2B_1-4A_0-8A_1x)\sin2x.\nonumber$$

Equating the coefficients of $x\cos2x$, $x\sin2x$, $\cos2x$, and $\sin2x$ here with the corresponding coefficients on the right side of Equation $\ref{eq:5.5.21}$ shows that $u_p$ is a solution of Equation $\ref{eq:5.5.21}$ if

$$\label{eq:5.5.22} \begin{array}{rcr} 8B_1&=-16\phantom{.}\\[4pt] -8A_1&=-\phantom{1}8\phantom{.}\\[4pt] \phantom{-}4B_0+2A_1&=6\phantom{.}\\[4pt] -4A_0+2B_1&=-8. \end{array}$$

The solution of this system is $A_1=1$, $B_1=-2$, $B_0=1$, $A_0=1$. Therefore

$$u_p=x[(1+x)\cos2x+(1-2x)\sin2x]\nonumber$$

is a particular solution of Equation $\ref{eq:5.5.21}$, and

$$y_p=xe^{-x}\left[(1+x)\cos2x+(1-2x)\sin2x\right]\nonumber$$

is a particular solution of Equation $\ref{eq:5.5.20}$.

You can also find a particular solution of Equation $\ref{eq:5.5.20}$ by substituting

$$y_p=xe^{-x}\left[(A_0+A_1x)\cos 2x +(B_0+B_1x)\sin 2x\right]\nonumber$$

for $y$ in Equation $\ref{eq:5.5.20}$ and equating the coefficients of $xe^{-x}\cos2x$, $xe^{-x}\sin2x$, $e^{-x}\cos2x$, and $e^{-x}\sin2x$ in the resulting expression for

$$y_p''+2y_p'+5y_p\nonumber$$

with the corresponding coefficients on the right side of Equation $\ref{eq:5.5.20}$. (See Exercise 5.5.38). This leads to the same system Equation $\ref{eq:5.5.22}$ of equations for $A_0$, $A_1$, $B_0$, and $B_1$ that we obtained in Example 5.5.6 . However, if you try this approach you’ll see that deriving Equation $\ref{eq:5.5.22}$ this way is much more tedious than the way we did it in Example 5.5.6 .