Differential Equations

Good God!

It's been a while, but i can help with a little bit of that.

L^(-1) {4s} = 4

L^(-1) {(3s-16)/(s^2+64)} =

L^(-1) { 3s / (s^s + 64) } + L^(-1) { -16 / (s^2 + 64) } =

L^(-1) {3 (s / (s^2 + 8^2) } + L^(-1) { 2 ( (8 / 8) / (s^2 + 8^2)) } =

3cos(8t) - 2sin(8t)

I believe that L^(-1) {1/(s-1)^2 } = t * exp(t) so the entire answer for your first question may be

4 + 3cos(8t) - 2sin(8t) + t*exp(t)

I don't have time for the second. There's a good chance i'm wrong with this, but i wanted to try.

I'll try the second one. Same caveat as fachatta

L^(-1) { (s+7) / (s^(2) + 6s + 25) } =L^(-1){(s+7)/((s+3)^2+4^2)}=
L^(-1)){(s+3)/((s+3)^2+4^2)+4/((s+3)^2+4^2)}
=exp(-3t)sin(4t)+exp(-3t)cos(4t)

Rap

I took the first Laplace transform

L^(-1) { (4/s) + (1 / (s-1)^2) + ( (3s-16) / (s^(2) + 64) ) }
and rearranged it to
L^(-1) { 4(1/s) + (1 / (s-1)^2) + 3(s / (s^(2) + 8^2)-2(8/(s^(2) + 8^2) )}
which becomes
4+t*exp(t)+3cos(8t)-2sin(8t)
which is the same thing fachatta got, so if we are mistaken we are mistaken in common.

Rap