Question

Problem 1: Use the Laplace Transforms to solve:


`y'' - y' = {e}^( - 3t) , \\ y(0) =y'(0) =0`
​

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)) = \frac1{s+3}`

`(s^2 - s) Y(s) = \frac1{s+3}`

`Y(s) = \frac1{(s^2 - s)(s+3)} = \frac1{s(s-1)(s+3)}`

Partial fractions:

`Y(s) = \frac as + \frac b{s-1} + \frac 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=-\frac13, b=\frac14, c=\frac1{12}`

`\implies Y(s) = -\frac13 * \frac1s + \frac14 * \frac1{s-1} + \frac1{12} * \frac1{s+3}`

Take the inverse transform:

`\boxed{y(t) = -\frac13 + \frac{e^t}4 + (e^(-3t))/(12)}`