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9.3: Undetermined Coefficients for Higher Order Equations

( \newcommand{\kernel}{\mathrm{null}\,}\)

In this section we consider the constant coefficient equation

a0y(n)+a1y(n1)++any=F(x),

where n3 and F is a linear combination of functions of the form

eαx(p0+p1x++pkxk)

or

eλx[(p0+p1x++pkxk)cosωx+(q0+q1x++qkxk)sinωx].

From Theorem 9.1.5, the general solution of Equation ??? is y=yp+yc, where yp is a particular solution of Equation ??? and yc is the general solution of the complementary equation

a0y(n)+a1y(n1)++any=0.

In Section 9.2 we learned how to find yc. Here we will learn how to find yp when the forcing function has the form stated above. The procedure that we use is a generalization of the method that we used in Sections 5.4 and 5.5, and is again called method of undetermined coefficients. Since the underlying ideas are the same as those in these section, we’ll give an informal presentation based on examples.

Forcing Functions of the Form eαx(p0+p1x+...+pkxk)

We first consider equations of the form

a0y(n)+a1y(n1)++any=eαx(p0+p1x++pkxk).

Example 9.3.1

Find a particular solution of

y

Solution

Substituting

\begin{align*} y&=ue^x,\\[4pt] y'&=e^x(u'+u),\\[4pt] y''&=e^x(u''+2u'+u),\\[4pt] y'''&=e^x(u'''+3u''+3u'+u)\end{align*}

into Equation \ref{eq:9.3.2} and canceling e^x yields

(u'''+3u''+3u'+u)+3(u''+2u'+u)+2(u'+u)-u =21+24x+28x^2+5x^3, \nonumber

or

\label{eq:9.3.3} u'''+6u''+11u'+5u=21+24x+28x^2+5x^3.

Since the unknown u appears on the left, we can see that Equation \ref{eq:9.3.3} has a particular solution of the form

u_p=A+Bx+Cx^2+Dx^3. \nonumber

Then

\begin{aligned} u_p'&=B+2Cx+3Dx^2\\[4pt] u_p''&=2C+6Dx\\[4pt] u_p'''&=6D.\end{aligned} \nonumber

Substituting from the last four equations into the left side of Equation \ref{eq:9.3.3} yields

\begin{aligned} u_p'''+6u_p''+11u_p'+5u_p&=6D+6(2C+6Dx)+11(B+2Cx+3Dx^2)\\[4pt] &\:\:+5(A+Bx+Cx^2+Dx^3)\\[4pt] &=(5A+11B+12C+6D)+(5B+22C+36D)x\\[4pt]&\:\:+(5C+33D)x^2+5Dx^3.\end{aligned} \nonumber

Comparing coefficients of like powers of x on the right sides of this equation and Equation \ref{eq:9.3.3} shows that u_p satisfies Equation \ref{eq:9.3.3} if

\begin{array}{rcr} 5D&=5\phantom{.}\\[4pt] 5C+33D&=28\phantom{.}\\[4pt] 5B+22C+36D&=24\phantom{.}\\[4pt] 5A+11B+12C+\phantom{3}6D&=21. \end{array}\nonumber

Solving these equations successively yields D=1, C=-1, B=2, A=1. Therefore

u_p=1+2x - x^2 + x^3 \nonumber

is a particular solution of Equation \ref{eq:9.3.3}, so

y_p = e^xu_p = e^x(1+2x-x^2+x^3) \nonumber

is a particular solution of Equation \ref{eq:9.3.2} (Figure 9.3.1 ).

fig090301.svg
Figure 9.3.1 : y_p=e^x(1+2x-x^2+x^3)
Example 9.3.2

Find a particular solution of

\label{eq:9.3.4} y^{(4)}-y'''-6y''+4y'+8y=e^{2x}(4+19x+6x^2).

Solution

Substituting

\begin{aligned} y&=ue^{2x},\\[4pt] y'&=e^{2x}(u'+2u),\\[4pt] y''&=e^{2x}(u''+4u'+4u),\\[4pt] y'''&=e^{2x}(u'''+6u''+12u'+8u),\\[4pt] y^{(4)}&=e^{2x}(u^{(4)}+8u'''+24u''+32u'+16u)\end{aligned} \nonumber

into Equation \ref{eq:9.3.4} and canceling e^{2x} yields

\begin{aligned} &&(u^{(4)}+8u'''+24u''+32u'+16u)-(u'''+6u''+12u'+8u)\\[4pt] &&-6(u''+4u'+4u)+4(u'+2u)+8u=4+19x+6x^2,\end{aligned} \nonumber

or

\label{eq:9.3.5} u^{(4)}+7u'''+12u''=4+19x+6x^2.

Since neither u nor u' appear on the left, we can see that Equation \ref{eq:9.3.5} has a particular solution of the form

\label{eq:9.3.6} u_p=Ax^2+Bx^3+Cx^4.

Then

\begin{aligned} u_p'&=2Ax+3Bx^2+4Cx^3\\[4pt] u_p''&=2A+6Bx+12Cx^2\\[4pt] u_p'''&=6B+24Cx\\[4pt] u_p^{(4)}&=24C.\end{aligned} \nonumber

Substituting u_p'', u_p''', and u_p^{(4)} into the left side of Equation \ref{eq:9.3.5} yields

\begin{aligned} u_p^{(4)}+7u_p'''+12u_p''&=24C+7(6B+24Cx)+12(2A+6Bx+12Cx^2)\\[4pt] &=(24A+42B+24C)+(72B+168C)x+144Cx^2.\end{aligned} \nonumber

Comparing coefficients of like powers of x on the right sides of this equation and Equation \ref{eq:9.3.5} shows that u_p satisfies Equation \ref{eq:9.3.5} if

\begin{array}{rrr} 144C&=6\phantom{.}\\[4pt] 72B+168C&=19\phantom{.}\\[4pt] 24A+42B+\phantom{1}24C&=4. \end{array}\nonumber

Solving these equations successively yields C=1/24, B=1/6, A=-1/6. Substituting these into Equation \ref{eq:9.3.6} shows that

u_p=\dfrac{x^2}{24}(-4+4x+x^2) \nonumber

is a particular solution of Equation \ref{eq:9.3.5}, so

y_p=e^{2x}u_p={x^2e^{2x}\over24}(-4+4x+x^2) \nonumber

is a particular solution of Equation \ref{eq:9.3.4} (Figure 9.3.2 ).

fig090302.svg
Figure 9.3.2 : y_p=\dfrac{x^2e^{2x}}{24}(-4+4x+x^2)

Forcing Functions of the Form e^{αx}(P(x) \cos ωx + Q(x) \sin ωx)

We now consider equations of the form

a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny= e^{\lambda x}\left(P(x)\cos\omega x+Q(x)\sin\omega x\right), \nonumber

where P and Q are polynomials.

Example 9.3.3

Find a particular solution of

\label{eq:9.3.7} y'''+y''-4y'-4y=e^x[(5-5x)\cos x+(2+5x)\sin x].

Solution

Substituting

\begin{aligned} y &=ue^x,\\[4pt] y' &=e^x(u'+u),\\[4pt] y'' &=e^x(u''+2u'+u),\\[4pt] y''' &=e^x(u'''+3u''+3u'+u)\end{aligned} \nonumber

into Equation \ref{eq:9.3.7} and canceling e^x yields

(u'''+3u''+3u'+u)+(u''+2u'+u)-4(u'+u)-4u =(5-5x)\cos x+(2+5x)\sin x,\nonumber

or

\label{eq:9.3.8} u'''+4u''+u'-6u=(5-5x)\cos x+(2+5x)\sin x.

Since \cos x and \sin x are not solutions of the complementary equation

u'''+4u''+u'-6u=0,\nonumber

a theorem analogous to Theorem 5.5.1 implies that Equation \ref{eq:9.3.8} has a particular solution of the form

\label{eq:9.3.9} u_p=(A_0+A_1x)\cos x+(B_0+B_1x)\sin x.

Then

\begin{align*} u_p' &=(A_1+B_0+B_1x)\cos x+(B_1-A_0-A_1x)\sin x,\\[4pt] u_p' &=(2B_1-A_0-A_1x)\cos x-(2A_1+B_0+B_1x)\sin x,\\[4pt] u_p''' &=-(3A_1+B_0+B_1x)\cos x-(3B_1-A_0-A_1x)\sin x,\end{align*}

so

u_p'''+4u_p''+u_p'-6u_p =-\left[10A_0+2A_1-8B_1+10A_1x\right]\cos x - \left[10B_0+2B_1+8A_1+10B_1x\right]\sin x. \nonumber

Comparing the coefficients of x\cos x, x\sin x, \cos x, and \sin x here with the corresponding coefficients in Equation \ref{eq:9.3.8} shows that u_p is a solution of Equation \ref{eq:9.3.8} if

\begin{align*} -10A_1 &=-5 \\[4pt] -10B_1&= 5\\[4pt] -10A_0-2A_1+8B_1&=5\\[4pt] -10B_0-2B_1-8A_1&=2. \end{align*}

Solving the first two equations yields A_1=1/2, B_1=-1/2. Substituting these into the last two equations yields

\begin{aligned} -10A_0&=5+2A_1-8B_1=10\\[4pt] -10B_0&=2+2B_1+8A_1=5, \end{aligned} \nonumber

so A_0=-1, B_0=-1/2. Substituting A_0=-1, A_1=1/2, B_0=-1/2, B_1=-1/2 into Equation \ref{eq:9.3.9} shows that

u_p=-\frac{1}{2}\left[(2-x)\cos x+(1+x)\sin x\right] \nonumber

is a particular solution of Equation \ref{eq:9.3.8}, so

y_p=e^xu_p=-{e^x\over2}\left[(2-x)\cos x+(1+x)\sin x\right] \nonumber

is a particular solution of Equation \ref{eq:9.3.7} (Figure 9.3.3 ).

fig090303.svg
Figure 9.3.3 : y_p=\{e^xu_p=-\dfrac{e^x}{2}\left[(2-x)\cos x+(1+x)\sin x\right]
Example 9.3.4

Find a particular solution of

\label{eq:9.3.10} y'''+4y''+6y'+4y= e^{-x}\left[(1-6x)\cos x-(3+2x)\sin x\right].

Solution

Substituting

\begin{aligned} y&=ue^{-x},\\[4pt] y'&=e^{-x}(u'-u),\\[4pt] y''&=e^{-x}(u''-2u'+u),\\[4pt] y'''&=e^{-x}(u'''-3u''+3u'-u)\end{aligned} \nonumber

into Equation \ref{eq:9.3.10} and canceling e^{-x} yields

(u'''-3u''+3u'-u)+4(u''-2u'+u)+6(u'-u)+4u =(1-6x)\cos x-(3+2x)\sin x, \nonumber

or

\label{eq:9.3.11} u'''+u''+u'+u=(1-6x)\cos x-(3+2x)\sin x.

Since \cos x and \sin x are solutions of the complementary equation

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

a theorem analogous to Theorem 5.5.1 implies that Equation \ref{eq:9.3.11} has a particular solution of the form

\label{eq:9.3.12} u_p=(A_0x+A_1x^2)\cos x+(B_0x+B_1x^2)\sin x.

Then

\begin{align*} u_p' =&[A_0+(2A_1+B_0)x+B_1x^2]\cos x+[B_0+(2B_1-A_0)x-A_1x^2]\sin x,\\[4pt] u_p'' =&[2A_1+2B_0-(A_0-4B_1)x-A_1x^2]\cos x\\[4pt]&+ [2B_1-2A_0-(B_0+4A_1)x-B_1x^2]\sin x,\\[4pt] u_p''' =& -[3A_0-6B_1+(6A_1+B_0)x+B_1x^2]\cos x \\[4pt]&-[3B_0+6A_1+(6B_1-A_0)x-A_1x^2]\sin x,\end{align*}

so

\begin{align*} u_p'''+u_p''+u_p'+u_p =& -[2A_0-2B_0-2A_1-6B_1+(4A_1-4B_1)x]\cos x\\[4pt] &-[2B_0+2A_0-2B_1+6A_1+(4B_1+4A_1)x]\sin x. \end{align*}

Comparing the coefficients of x\cos x, x\sin x, \cos x, and \sin x here with the corresponding coefficients in Equation \ref{eq:9.3.11} shows that u_p is a solution of Equation \ref{eq:9.3.11} if

\begin{array}{rcr} -4A_1+4B_1&=-6\phantom{.}\\[4pt] -4A_1-4B_1&=-2\phantom{.}\\[4pt] -2A_0+2B_0+2A_1+6B_1&=\phantom{-}1\phantom{.}\\[4pt] -2A_0-2B_0-6A_1+2B_1&=-3. \end{array}\nonumber

Solving the first two equations yields A_1=1, B_1=-1/2. Substituting these into the last two equations yields

\begin{aligned} -2A_0+2B_0&=\phantom{-}1-2A_1-6B_1=2\phantom{.}\\[4pt] -2A_0-2B_0&=-3+6A_1-2B_1=4, \end{aligned} \nonumber

fig090304.svg
Figure 9.3.4 : y_p=-\dfrac{xe^{-x}}{2}\left[(3-2x)\cos x+(1+x)\sin x\right]

so A_0=-3/2 and B_0=-1/2. Substituting A_0=-3/2, A_1=1, B_0=-1/2, B_1=-1/2 into Equation \ref{eq:9.3.12} shows that

u_p=- \dfrac{x}{2}\left[(3-2x)\cos x+(1+x)\sin x\right] \nonumber

is a particular solution of Equation \ref{eq:9.3.11}, so

y_p=e^{-x}u_p=-{xe^{-x}\over2}\left[(3-2x)\cos x+(1+x)\sin x\right] \nonumber

Figure 9.3.4 is a particular solution of Equation \ref{eq:9.3.10}.


This page titled 9.3: Undetermined Coefficients for Higher Order Equations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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