Chapter 17.2, Problem 7E

### 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.7. y" – 2y' + 5y = sin x, y(0) = 1, y'(0) = 1

To determine

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

Explanation

Given data:

The differential equation is,

yâ€³+2yâ€²+5y=sinx (1)

with y(0)=1,â€‰â€‰andâ€‰â€‰yâ€²(0)=1 .

Consider the auxiliary equation is,

r2âˆ’2r+5=0 (2)

Roots of equation (2) are,

r=2Â±(âˆ’2)2âˆ’4(1)(5)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€‰}=1Â±2i

Write the expression for the complementary solution,

yc(x)=ex(c1cos2x+c2sin2x)

The particular solution yp(x) is,

yp(x)=Acosx+Bsinx (3)

Differentiate equation (3) with respect to x,

yâ€²p(x)=ddx(Acosx+Bsinx)

yâ€²p(x)=âˆ’Asinx+Bcosx (4)

Differentiate equation (3) with respect to x,

yâ€³p(x)=ddx(âˆ’Asinx+Bcosx)

yâ€³p(x)=âˆ’Acosxâˆ’Bsinx (5)

Substitute equations (3), (4) and (5) in equation (1),

âˆ’Acosxâˆ’Bsinx+2(âˆ’Asinx+Bcosx)+5(Acosx+Bsinx)=sinx

(4Aâˆ’2B)cosx+(2A+4B)sinx=sinx (6)

Substitute 0 for x in equation (6),

(4Aâˆ’2B)cos0+(2A+4B)sin0=sin0(4Aâˆ’2B)(1)+0=0

4Aâˆ’2B=0 (7)

Substitute 90Â° for x in equation (6),

(4Aâˆ’2B)cos90Â°+(2A+4B)sin90Â°=sin90Â°(4Aâˆ’2B)(0)+(2A+4B)(1)=(1)

2A+4B=1 (8)

Multiply 2 on both sides of equation (8),

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