Chapter 17.2, Problem 28E

### 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 using the method of variation of parameters.28. y ″ + 4 y ′ + 4 y = e − 2 x x 3

To determine

To solve: The differential equation by using method of variation of parameters.

Explanation

Given data:

The differential equation is,

yâ€³+4yâ€²+4y=eâˆ’2xx3 (1)

Consider the auxiliary equation is,

r2+4r+4=0 (2)

Roots of equation (2) are,

r=âˆ’(4)Â±(4)2âˆ’4(1)(4)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’42=âˆ’2

Write the expression for the complementary solution of the one real root.

yc(x)=c1erx+c2xerx (3)

Substitute âˆ’2 for r in equation (3),

yc(x)=c1eâˆ’2x+c2xeâˆ’2x (4)

From equation (4), set y1=eâˆ’2x and y2=xeâˆ’2x .

Calculate y1yâ€²2âˆ’y2yâ€²1 .

y1yâ€²2âˆ’y2yâ€²1=eâˆ’2xd(xeâˆ’2x)dxâˆ’xeâˆ’2xd(eâˆ’2x)dx=eâˆ’2x(xeâˆ’2x(âˆ’2)+eâˆ’2x(1))âˆ’xeâˆ’2xeâˆ’2x(âˆ’2)=eâˆ’2x(âˆ’2x+1)eâˆ’2x+2xeâˆ’2xeâˆ’2x=âˆ’2xeâˆ’2xeâˆ’2x+eâˆ’2xeâˆ’2x+2xeâˆ’2xeâˆ’2x

Simplify the equation.

y1yâ€²2âˆ’y2yâ€²1=eâˆ’2xeâˆ’2x=eâˆ’4x

Write the expression to find the arbitrary function uâ€²1 ,

uâ€²1=âˆ’G(x)y2y1yâ€²2âˆ’y2yâ€²1

Here,

G(x) is the expression for R.H.S of differential equation in (1),

Substitute eâˆ’2xx3 for G(x) , xeâˆ’2x for y2 , and eâˆ’4x for y1yâ€²2âˆ’y2yâ€²1 ,

uâ€²1=âˆ’eâˆ’2xx3(xeâˆ’2x)eâˆ’4x=âˆ’eâˆ’2x(xeâˆ’2x)x3eâˆ’4x=âˆ’1x2

Integrate on both sides of the equation

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