Solving $y''+4y=g(t)$, $y(0)=3$, $y'(0)=-1$ using Laplace Transforms.
I get $Y(s)=\frac{G(s)}{s^{2}+4}+\frac{3s}{s^{2}+4}-\frac{1}{s^2+4}$
Then using Inverse Laplace Transforms, I am not sure of my result: I consider $\frac{G(s)}{s^{2}+4}$ as the Laplace Transform of the convolution $g(t)$ and $f(t)=sin(2t)$ which gives me:
$y(t)=\frac{1}{2} (sin(2t)\ast g(t))+3cos(2t)-\frac{1}{2}sin(2t)=\frac{1}{2} \int_{0}^t sin(\tau)g(t-\tau)d\tau +3cos(2t)-\frac{1}{2}sin(2t)=\frac{1}{2} \int_{0}^t g(\tau)sin(2t-\tau)d\tau +3cos(2t)-\frac{1}{2}sin(2t)$
Since the convolution is commutative, I write the last equality twice, is this the way to commute the convolution? Also, is this the solution of the IVP?