9.3: Undetermined Coefficients for Higher Order Equations
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( \newcommand{\kernel}{\mathrm{null}\,}\)
In this section we consider the constant coefficient equation
a0y(n)+a1y(n−1)+⋯+any=F(x),
where n≥3 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(n−1)+⋯+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(n−1)+⋯+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 ).
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 ).
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 ).
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
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}.