9.3: Undetermined Coefficients for Higher Order Equations
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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).
Find a particular solution of
y‴+3y″+2y′−y=ex(21+24x+28x2+5x3).
Solution
Substituting
y=uex,y′=ex(u′+u),y″=ex(u″+2u′+u),y‴=ex(u‴+3u″+3u′+u)
into Equation ??? and canceling ex yields
(u‴+3u″+3u′+u)+3(u″+2u′+u)+2(u′+u)−u=21+24x+28x2+5x3,
or
u‴+6u″+11u′+5u=21+24x+28x2+5x3.
Since the unknown u appears on the left, we can see that Equation ??? has a particular solution of the form
up=A+Bx+Cx2+Dx3.
Then
u′p=B+2Cx+3Dx2u″p=2C+6Dxu‴p=6D.
Substituting from the last four equations into the left side of Equation ??? yields
u‴p+6u″p+11u′p+5up=6D+6(2C+6Dx)+11(B+2Cx+3Dx2)+5(A+Bx+Cx2+Dx3)=(5A+11B+12C+6D)+(5B+22C+36D)x+(5C+33D)x2+5Dx3.
Comparing coefficients of like powers of x on the right sides of this equation and Equation ??? shows that up satisfies Equation ??? if
5D=5.5C+33D=28.5B+22C+36D=24.5A+11B+12C+36D=21.
Solving these equations successively yields D=1, C=−1, B=2, A=1. Therefore
up=1+2x−x2+x3
is a particular solution of Equation ???, so
yp=exup=ex(1+2x−x2+x3)
is a particular solution of Equation ??? (Figure 9.3.1 ).
Find a particular solution of
y(4)−y‴−6y″+4y′+8y=e2x(4+19x+6x2).
Solution
Substituting
y=ue2x,y′=e2x(u′+2u),y″=e2x(u″+4u′+4u),y‴=e2x(u‴+6u″+12u′+8u),y(4)=e2x(u(4)+8u‴+24u″+32u′+16u)
into Equation ??? and canceling e2x yields
(u(4)+8u‴+24u″+32u′+16u)−(u‴+6u″+12u′+8u)−6(u″+4u′+4u)+4(u′+2u)+8u=4+19x+6x2,
or
u(4)+7u‴+12u″=4+19x+6x2.
Since neither u nor u′ appear on the left, we can see that Equation ??? has a particular solution of the form
up=Ax2+Bx3+Cx4.
Then
u′p=2Ax+3Bx2+4Cx3u″p=2A+6Bx+12Cx2u‴p=6B+24Cxu(4)p=24C.
Substituting u″p, u‴p, and u(4)p into the left side of Equation ??? yields
u(4)p+7u‴p+12u″p=24C+7(6B+24Cx)+12(2A+6Bx+12Cx2)=(24A+42B+24C)+(72B+168C)x+144Cx2.
Comparing coefficients of like powers of x on the right sides of this equation and Equation ??? shows that up satisfies Equation ??? if
144C=6.72B+168C=19.24A+42B+124C=4.
Solving these equations successively yields C=1/24, B=1/6, A=−1/6. Substituting these into Equation ??? shows that
up=x224(−4+4x+x2)
is a particular solution of Equation ???, so
yp=e2xup=x2e2x24(−4+4x+x2)
is a particular solution of Equation ??? (Figure 9.3.2 ).
Forcing Functions of the Form eαx(P(x)cosωx+Q(x)sinωx)
We now consider equations of the form
a0y(n)+a1y(n−1)+⋯+any=eλx(P(x)cosωx+Q(x)sinωx),
where P and Q are polynomials.
Find a particular solution of
y‴+y″−4y′−4y=ex[(5−5x)cosx+(2+5x)sinx].
Solution
Substituting
y=uex,y′=ex(u′+u),y″=ex(u″+2u′+u),y‴=ex(u‴+3u″+3u′+u)
into Equation ??? and canceling ex yields
(u‴+3u″+3u′+u)+(u″+2u′+u)−4(u′+u)−4u=(5−5x)cosx+(2+5x)sinx,
or
u‴+4u″+u′−6u=(5−5x)cosx+(2+5x)sinx.
Since cosx and sinx are not solutions of the complementary equation
u‴+4u″+u′−6u=0,
a theorem analogous to Theorem 5.5.1 implies that Equation ??? has a particular solution of the form
up=(A0+A1x)cosx+(B0+B1x)sinx.
Then
u′p=(A1+B0+B1x)cosx+(B1−A0−A1x)sinx,u′p=(2B1−A0−A1x)cosx−(2A1+B0+B1x)sinx,u‴p=−(3A1+B0+B1x)cosx−(3B1−A0−A1x)sinx,
so
u‴p+4u″p+u′p−6up=−[10A0+2A1−8B1+10A1x]cosx−[10B0+2B1+8A1+10B1x]sinx.
Comparing the coefficients of xcosx, xsinx, cosx, and sinx here with the corresponding coefficients in Equation ??? shows that up is a solution of Equation ??? if
−10A1=−5−10B1=5−10A0−2A1+8B1=5−10B0−2B1−8A1=2.
Solving the first two equations yields A1=1/2, B1=−1/2. Substituting these into the last two equations yields
−10A0=5+2A1−8B1=10−10B0=2+2B1+8A1=5,
so A0=−1, B0=−1/2. Substituting A0=−1, A1=1/2, B0=−1/2, B1=−1/2 into Equation ??? shows that
up=−12[(2−x)cosx+(1+x)sinx]
is a particular solution of Equation ???, so
yp=exup=−ex2[(2−x)cosx+(1+x)sinx]
is a particular solution of Equation ??? (Figure 9.3.3 ).
Find a particular solution of
y‴+4y″+6y′+4y=e−x[(1−6x)cosx−(3+2x)sinx].
Solution
Substituting
y=ue−x,y′=e−x(u′−u),y″=e−x(u″−2u′+u),y‴=e−x(u‴−3u″+3u′−u)
into Equation ??? and canceling e−x yields
(u‴−3u″+3u′−u)+4(u″−2u′+u)+6(u′−u)+4u=(1−6x)cosx−(3+2x)sinx,
or
u‴+u″+u′+u=(1−6x)cosx−(3+2x)sinx.
Since cosx and sinx are solutions of the complementary equation
u‴+u″+u′+u=0,
a theorem analogous to Theorem 5.5.1 implies that Equation ??? has a particular solution of the form
up=(A0x+A1x2)cosx+(B0x+B1x2)sinx.
Then
u′p=[A0+(2A1+B0)x+B1x2]cosx+[B0+(2B1−A0)x−A1x2]sinx,u″p=[2A1+2B0−(A0−4B1)x−A1x2]cosx+[2B1−2A0−(B0+4A1)x−B1x2]sinx,u‴p=−[3A0−6B1+(6A1+B0)x+B1x2]cosx−[3B0+6A1+(6B1−A0)x−A1x2]sinx,
so
u‴p+u″p+u′p+up=−[2A0−2B0−2A1−6B1+(4A1−4B1)x]cosx−[2B0+2A0−2B1+6A1+(4B1+4A1)x]sinx.
Comparing the coefficients of xcosx, xsinx, cosx, and sinx here with the corresponding coefficients in Equation ??? shows that up is a solution of Equation ??? if
−4A1+4B1=−6.−4A1−4B1=−2.−2A0+2B0+2A1+6B1=−1.−2A0−2B0−6A1+2B1=−3.
Solving the first two equations yields A1=1, B1=−1/2. Substituting these into the last two equations yields
−2A0+2B0=−1−2A1−6B1=2.−2A0−2B0=−3+6A1−2B1=4,
so A0=−3/2 and B0=−1/2. Substituting A0=−3/2, A1=1, B0=−1/2, B1=−1/2 into Equation ??? shows that
up=−x2[(3−2x)cosx+(1+x)sinx]
is a particular solution of Equation ???, so
yp=e−xup=−xe−x2[(3−2x)cosx+(1+x)sinx]
Figure 9.3.4 is a particular solution of Equation ???.