solve the initial value problem y9y20et y00 y01SolutionFirs

solve the initial value problem: y\"+9y=20e^-t ; y(0)=0; y\'(0)=1

Solution

First we solve the homogeneous equation:

y\'\'=-3^2 y

The general solution is:

yh = a sin(3t) +b cos(3t)

Let guess for the particular solution be:

A e^{-t}

Substituting gives

-A e^{-t}+9A e^{-t}=20 e^{-t}

8 A =20

A=5/2

So,

y(t)= a sin(3t) +b cos(3t) +5 e^{-t}/2

y(0)=0

Hence,

b+5/2=0 ie b=-5/2

y\'(0)=1

3a-5/2=1

3a=7/2

a=7/6