Compute the solution to the following initial value problem

Solution

y \' \' +9y=cos(t)

for particular integral:

let y=Acos(t)+Bsin(t)

- Acost - Bsint +9Acost+9Bsint= cost

comparing the values we get

A=1/8 ,B=0

y_p =(1/8)cos(t)

for complementary function

(D^2 +9)y=0

since roots of characteristic equation are imagenary i.e +3i,-3i

y_comp =Ccos(3t)+Dsin(3t)

from initial conditions : y(0)=0=y \' (0) =>y \' \'(0) =1

C= -(1/9) , D=0

so, general solution is : y= (-1/9)cos(3t) +(1/8)cos(t) .