## Find the Inverse Laplace Transforms

Find the inverse Laplace transform of: 3s+5s(s2+9) Workings: 3s+5s(s2+9) =3ss(s2+9+5s(s2+9) =3s2+9+5s1s2+9 =sin(3t)+5s1s2+9 Now I’m not to sure on what to do. Any help will be appreciated. Answer The convolution theorem is, found here, L−1{f(s)g(s)}=∫t0f(t−u)g(u)du In the case here f(s)=1/s which is the transform of 1 and g(s) being the transform of sin. Now L−1{1s(s2+a2)}=1a∫t0(1)sin(au)du=−1a[cos(au)a]t0=−1a2(cos(at)−1) With … Read more