Chapter 17.2, Problem 9E

### 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.9. y" –y' = xex, y(0) = 2, y'(0) = 1

To determine

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

Explanation

Given data:

The differential equation is,

yâ€³âˆ’yâ€²=xex (1)

with y(0)=2â€‰â€‰andâ€‰â€‰yâ€²(0)=1 .

Consider the auxiliary equation is,

r2âˆ’r=0 (2)

Roots of equation (2) are,

r=0Â±(0)2âˆ’4(1)(âˆ’1)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€‰}=0â€‰andâ€‰1

Write the expression for the complementary solution,

yc(x)=c1+c2ex

The particular solution yp(x) is,

yp(x)=x(Ax+B)ex

yp(x)=(Ax2+Bx)ex (3)

Differentiate equation (3) with respect to x,

yâ€²p(x)=ddx((Ax2+Bx)ex)

yâ€²p(x)=(Ax2+(2A+B)x+B)ex (4)

Differentiate equation (3) with respect to x,

yâ€³p(x)=ddx((Ax2+(2A+B)x+B)ex)

yâ€³p(x)=(Ax2+(4A+B)x+(2A+2B))ex (5)

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

(Ax2+(4A+B)x+(2A+2B))exâˆ’(Ax2+(2A+B)x+B)ex=xex

(2Ax+(2A+B))ex=xex

2Ax+(2A+B)=x (6)

Differentiate the equation (6) with respect to x,

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

Simplify the equation

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