Question

Problem 1: Use the Laplace Transforms to solve:

Answer 1

Transforming the ODE yields

$L\left\{y'' - y'\right\} = L\left\{e^{-3t}\right\}$

$(s^2 Y(s) - sy(0) - y'(0)) - (s Y(s) - y(0)) = \dfrac1{s+3}$

$(s^2 - s) Y(s) = \dfrac1{s+3}$

$Y(s) = \dfrac1{(s^2 - s)(s+3)} = \dfrac1{s(s-1)(s+3)}$

Partial fractions:

$Y(s) = \dfrac as + \dfrac b{s-1} + \dfrac c{s+3}$

$\implies 1 = a(s-1)(s+3) + b s(s+3) + c s(s - 1)$

$\implies 1 = -3 a + (2 a + 3 b - c) s + (a + b + c) s^2$

$\implies \begin{cases}-3a=1 \\ 2a+3b-c=0 \\ a+b+c=0\end{cases} \implies a=-\dfrac13, b=\dfrac14, c=\dfrac1{12}$

$\implies Y(s) = -\dfrac13 \times \dfrac1s + \dfrac14 \times \dfrac1{s-1} + \dfrac1{12} \times \dfrac1{s+3}$

Take the inverse transform:

$\boxed{y(t) = -\dfrac13 + \dfrac{e^t}4 + \dfrac{e^{-3t}}{12}}$