   Chapter 17.2, Problem 6E

Chapter
Section
Textbook Problem

Solve the differential equation or initial-value problem using the method of undetermined coefficients.6. y" – 4y' + 4y = x – sin x

To determine

To solve: The differential equation by the method of undetermined coefficients.

Explanation

Given data:

The differential equation is,

y4y+4y=xsinx (1)

Consider the auxiliary equation is,

r24r+4=0 (2)

Roots of equation (2) are,

r=4±(4)24(1)(4)2(1){r=b±b24ac2afortheequationofar2+br+c=0}=4±02=2

Write the expression for the complementary solution,

yc(x)=c1e2x+c2xe2x

Re-write equation (1) by neglecting the term sinx ,

y4y+4y=x (3)

The particular solution yp1(x) is,

yp1(x)=Ax+B (4)

Differentiate equation (4) with respect to x,

yp1(x)=ddx(Ax+B)

yp1(x)=A (5)

Differentiate equation (5) with respect to x,

yp1(x)=ddx(A)

yp1(x)=0 (6)

Substitute equations (4), (5) and (6) in (3),

04A+4(Ax+B)=x

4Ax+(4B4A)=x (7)

Differentiate the equation (7) with respect to x,

ddx(4Ax+4B4A)=ddx(x)ddx(4Ax)+ddx(4B)+ddx(4A)=ddx(x)4A+0+0=14A=1

Simplify the equation,

A=14

Substitute 0 for x in equation (7),

4A(0)+(4B4A)=0

4B4A=0 (8)

Substitute 14 for A in equation (8),

4B4(14)=04B1=04B=1B=14

Substitute 14 for A and 14 for B in equation (4),

yp1(x)=14x+14 (9)

Re-write equation (1) by neglecting x.

y4y+4y=sinx (10)

The particular solution yp2(x) is,

yp2(x)=Acosx+Bsinx (11)

Differentiate equation (11) with respect to x,

yp2(x)=ddx(Acosx+Bsinx)

yp2(x)=Asinx+Bcosx (12)

Differentiate equation (12) with respect to x,

yp2(x)=ddx(Asinx+Bcosx)

yp2(x

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