i am trying to find the solution to this second order differential equation below.
$y''(t) + 4 \cdot y(t) = 0.5,$ with $y(0) = 2, y'(0) = -1$
i have made some progress on this.
$Y(s)$ is given by:
$L(y''(t)) + 4 \cdot L(y(t)) = 0.5 \cdot L(1)$
by the rules of laplace transforms:
$s^2 \cdot Y(s) - s \cdot y(0) - y'(0) + 4 \cdot Y(s) = s^2 * Y(s) - 2s + 1 + 4 \cdot Y(s)$
$s^2 \cdot Y(s) - 2s + + 4 \cdot Y(s) = \frac{1}{2s}$
$Y(s) \cdot (s^2 + 4) = \frac{1}{2s} + 2s - 1$
$Y(s) = \frac{1}{2s \cdot (s^2 + 4)} + \frac{2s}{s^2 + 4} - \frac{1}{s^2 + 4}$
this is where i got stuck. i don't know what to do with the $\frac{1}{2s \cdot (s^2 + 4)}$ term. i was thinking of using partial fractions, but i am not sure how to separate the original fraction.
any advice on how to proceed? thank you in advance.
You can separate using partial fractions. You just need to look for fractions of the form
$$\frac{1}{2s(s^2+4)}= \frac{A}{s}+\frac{Bs+C}{s^2+4}$$
By clearing denominators and equating coefficients of $s$ we see that $A= 1/8$, $B = -1/8$ and $C = 0$