Chapter 17, Problem 9RE

### 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.9. d 2 y d x 2 − d y d x − 6 y = 1 + e − 2 x

To determine

To solve: The differential equation.

Explanation

Given data:

The differential equation is,

d2ydx2âˆ’dydxâˆ’6y=1+eâˆ’2x

yâ€³âˆ’yâ€²âˆ’6y=1+eâˆ’2x (1)

Consider the auxiliary equation.

r2âˆ’râˆ’6=0 (2)

Roots of equation (2) are,

r=âˆ’(âˆ’1)Â±(âˆ’1)2âˆ’4(1)(âˆ’6)2(1)â€‰â€‰â€‰â€‰â€‰â€‰{âˆµr=âˆ’bÂ±b2âˆ’4ac2aforâ€‰theâ€‰equationâ€‰ofar2+br+c=0â€‰â€‰}=1Â±52=âˆ’2â€‰andâ€‰3

Write the expression for the complementary solution for two real roots,

yc(x)=c1er1x+c2er2x

Substitute âˆ’2 for r1 and 3 for r2 ,

yc(x)=c1eâˆ’2x+c2e3x (3)

Re-arrange equation (1) by neglecting eâˆ’2x ,

yâ€³âˆ’yâ€²âˆ’6y=1 (4)

Find the Particular solution for equation (4),

The Right hand side (RHS) of a differential equation contains only numerical value. Then, the particular solution for this case can be expressed as follows.

yp1(x)=A (5)

Differentiate equation (5) with respect to x.

yâ€²p1(x)=ddx(0)

yâ€²p1(x)=0 (6)

Differentiate equation (6) with respect to x.

yâ€³p1(x)=ddx(0)

yâ€³p1(x)=0 (7)

Substitute equations (5), (6), and (7) in equation (4),

0âˆ’0âˆ’6A=1A=âˆ’16

Re-arrange equation (1) by neglecting 1,

yâ€³âˆ’yâ€²âˆ’6y=eâˆ’2x (8)

Find the Particular solution for equation (8)

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