Consider the differential equation: y′′-8y′=7x+1. Find the general solution to the corresponding homogeneous equation. In your answer, use c1 and c2 to denote arbitrary constants. Enter c1 as c1 and c2 as c2. yc = Apply the method of undetermined coefficients to find a particular solution. yp=

yp = - x/8

Step-by-step explanation:

Given the differential equation: y′′-8y′=7x+1,

The solution of the DE will be the sum of the complementary solution (yc) and the particular integral (yp)

First we will calculate the complimentary solution by solving the homogenous part of the DE first i. e by equating the DE to zero and solving to have;

y′′-8y′=0

The auxiliary equation will give us;

m²-8m = 0

m (m-8) = 0

m = 0 and m-8 = 0

m1 = 0 and m2 = 8

Since the value of the roots are real and different, the complementary solution (yc) will give us

yc = Ae^m1x + Be^m2x

yc = Ae^0+Be^8x

yc = A+Be^8x

To get yp we will differentiate yc twice and substitute the answers into the original DE

yp = Ax+B (using the method of undetermined coefficients

y'p = A

y"p = 0

Substituting the differentials into the general DE to get the constants we have;

0-8A = 7x+1

Comparing coefficients

-8A = 1

A = - 1/8

B = 0

yp = - 1/8x+0

yp = - x/8 (particular integral)

y = yc+yp

y = A+Be^8x-x/8