Chapter 17.2, Problem 10E

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Solve the differential equation or initial-value problem using the method of undetermined coefficients.10. y" + y' – 2y = x + sin 2x, y(0) = 1, y'(0) = 0

To determine

To solve: The differential equation or initial-value problem by the method of undetermined coefficients.

Explanation

Given data:

The differential equation is,

yâ€³+yâ€²âˆ’2y=x+sin2x (1)

with y(0)=1â€‰â€‰andâ€‰â€‰yâ€²(0)=0 .

Consider the auxiliary equation is,

r2+râˆ’2=0 (2)

Roots of equation (2) are,

r=âˆ’1Â±(âˆ’1)2âˆ’4(1)(âˆ’2)2(1)â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰{âˆµr=âˆ’bÂ±b2âˆ’4ac2aforâ€‰theâ€‰equationâ€‰ofar2+br+c=0â€‰â€‰}=âˆ’1Â±32=1â€‰andâ€‰âˆ’2

Write the expression for the complementary solution,

yc(x)=c1er1x+c2er2x

Substitute 1 for r1 and âˆ’2 for r2 ,

yc(x)=c1ex+c2eâˆ’2x

The particular solution yp1(x) is,

yp1(x)=Ax+B (3)

Differentiate equation (3) with respect to x,

yâ€²p1(x)=ddx(Ax+B)

yâ€²p1(x)=A (4)

Differentiate equation (4) with respect to x,

yâ€³p1(x)=ddx(A)

yâ€³p1(x)=0 (5)

Re-write equation (1) by neglecting the term sin2x ,

yâ€³+yâ€²âˆ’2y=x (6)

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

0+Aâˆ’2(Ax+B)=x (7)

Differentiate the equation (7) with respect to x,

ddx(Aâˆ’2(Ax+B))=ddx(x)ddx(A)+ddx(âˆ’2Ax)+ddx(âˆ’2B)=ddx(x)0âˆ’2A+0=1âˆ’2A=1

Simplify the equation,

A=âˆ’12

Substitute 0 for x in equation (7),

0+Aâˆ’2(A(0)+B)=0

Aâˆ’2B=0 (8)

Substitute âˆ’12 for A in equation (8),

(âˆ’12)âˆ’2B=0âˆ’2B=12B=âˆ’14

Substitute âˆ’12 for A and âˆ’14 for B in equation (3),

yp1(x)=âˆ’12xâˆ’14 (9)

Re-write equation (1) by neglecting x.

yâ€³+yâ€²âˆ’2y=sin2x (10)

The particular solution yp2(x) is,

yp2(x)=Acos2x+Bsin2x (11)

Differentiate equation (11) with respect to x,

yâ€²p2(x)=ddx(Acos2x+Bsin2x)

yâ€²p2(x)=âˆ’2Asin2x+2Bcos2x (12)

Differentiate equation (12) with respect to x,

yâ€³p2(x)=ddx(âˆ’2Asin2x+2Bcos2x)

yâ€³p2(x)=âˆ’4Acos2xâˆ’4Bsin2x (13)

Substitute equations (11), (12), and (13) in equation (10),

(âˆ’4Acos2xâˆ’4Bsin2x)+(âˆ’2Asin2x+2Bcos2x)âˆ’2(Acos2x+Bsin2x)=sin2x(âˆ’4A+2Bâˆ’2A)cos2x+(âˆ’4Bâˆ’2Aâˆ’2B)sin2x=sin2x

(âˆ’6A+2B)cos2x+(âˆ’6Bâˆ’2A)sin2x=sin2x (14)

Substitute 0 for x in equation (14)

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