Solving an integral using Laplace transform and inverse Laplace transform

Question

I want to solve this integral equation using Laplace:

$$ Y(t) + 3{\int\limits_0^t Y(t)}\operatorname d\!t = 2cos(2t)$$

if $$ \mathcal{L}\{Y(t)\} = f(s)$$

then, $$ f(s) + 3 \frac{f(s)}{s} = \frac{2s}{s^2+4} $$

doing some operations I obtain

$$ f(s) = \frac{2s^2}{s^3+3s^2+4s+12} $$

and using Ruffini in the denominator

$$ f(s) = \frac{2s^2}{(s+3)(s^2+4)} $$

How can I calculate the inverse Laplace transform of the following function to get the solution?

Answer 1

Building on my comment, try the following: $$f(s)=\frac{2s^2}{(s+3)(s^2+4)} = \frac{As+B}{s^2+4} + \frac{C}{s+3}$$ So $As(s+3) + B(s+3) + C(s^2+4) = 2s^2$, whence\ \begin{align*} A+C &= 2\\ 3A + B &= 0\\ 3B+4C &= 0 \end{align*} We can shortcut the solution a bit, as follows:

Plug in $s = -3$, we get: $13C = 18$, whence $C = \frac{18}{13}$.

Plug in $s = 0$, we get $3B+4C = 0$, so $B = -\frac{24}{13}$.

Then using the first equation, we get $A = \frac{8}{13}$.

So we have

\begin{align*} f(s) &= \frac{\frac{8}{13}s}{s^2+4} - \frac{\frac{24}{13}}{s^2+4} + \frac{\frac{18}{13}}{s+3}\\ &= \frac{1}{13}\left(8\frac{s}{s^2+4} - 12\frac{2}{s^2+4} + \frac{18}{s+3}\right) \end{align*} And at this point I can refer you to a table of Laplace transforms (which implicitly gives the termwise inverse Laplace transform).

$$f(t) = \frac{1}{13}\left(8\cos(2t) - 12\sin(2t) + 18e^{-3t}\right)$$

Answer 2

Using partial fractions, we can obtain $$ \begin{align} \frac{2s^2}{(s+3)(s^2+4)}&=\frac{8(s-3)}{13(s^2+4)}+\frac{18}{13(s+3)}\\ &=\frac1{13}\left(\frac{8s}{s^2+2^2}-\frac{12\cdot2}{s^2+2^2}+\frac{18}{s-(-3)}\right). \end{align} $$ Take the inverse we will obtain $$ \frac1{13}\left(8\cos2t-12\sin2t+18\ e^{\large-3t}\right). $$