y8y16y2x3 y01 y00SolutionFirst we solve the homogeneous equa

y\'\'-8y\'+16y=2x-3, y(0)=1, y\'(0)=0

Solution

First we solve the homogeneous equation

y\'\'-8y\'+16y=0

Assume: y=e^{kx}, substituting gives

k^2-8k+16=0

k=4

So we have repeated roots so general solution to homogeneous equation is

yc(x)=e^{4x}(A+Bx)

Let particular solution guess based on inhomogeneous part be: Ax+B

Substituting gives

-8A+16Ax+16B=2x-3

So, 16A=2 ,

A=1/8

-8A+16B=-3

B=-1/8

So, general solution is:

y(x)=e^{4x}(A+Bx)-(1+x)/8