Laplace Transforms

Question

Solve the initial value problem for y(t) using Laplace Transforms.

$$L\{y''+3y'\}=L\{f(t)\}$$

$$s^2Y-sy(0)-sy'(0)+3(sY-sy(0))=L\{t\}+L\{1\}-L\{u4(t)(t-4)\}-5L{u8(t)}$$

$$Y(s^2+3s)=(1/s^2)+(1/s)-(1/s^2)e^{-4s}-5(1/s)e^{-8s}$$

Answer 1

You can write ${1 \over s(s^2+3s)} = {1 \over 9(s+3)} - { 1 \over 9s} + {1 \over 3 s^2}$ and ${1 \over s^2(s^2+3s)} = -{1 \over 27(s+3)} + { 1 \over 27s} - {1 \over 9 s^2}+ {1 \over 3 s^3}$.