Question

y''+9y=cos(2t) where y=y(t) , t>=0

y(0)=1 ; y(-pi/2)=3 ;
solve the differential equation using laplace transforms.
So how do i find y'(0) ?

Answer 1

\[l(f'(t))=s \times F(s) - f(0)\]

Answer 2

um i got no idea what you wrote there :|