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 .