Laplace transform of $\ddot{x} +4x = f(t)$ I am stuck in an excercise that on first sight didn't look that strange:
Given the initial problem:
$$\ddot{x} +4x = f(t), x(t=0)=3, \dot{x}(t=0)=-1$$
So I started:
$$s^2(X(s) -sx(0)-\dot{x}(0) +4X(s) = \mathcal{L}(f(t))$$
Now substitute the given values:
$$s^2X(s) -3s-(-1) +4X(s) = \mathcal{L}(f(t))$$
Rearranging:
$$X(s)(s^2+4) -3s+1 =  \mathcal{L}(f(t))$$
The answer give: The Laplace transform is of the form:
$$X(t)= A \cos2t+B \sin2t +\frac{1}{2}\int_{0}^{t}f(\tau)\sin2(t-\tau)d\tau $$
Is there anybody that can help me to get the given form?
 A: Hint
You were almost done put all the s terme at the right side then use the convolution formula 
$$X(s)(s^2+4) -3s+1 =  \mathcal{L}(f(t))$$
$$X(s)(s^2+4) =3s-1+  \mathcal{L}(f(t))$$
For convenience I substitute $h(s)=\mathcal{L}(f(t))$
$$X(s) =\frac {3s-1}{s^2+4}+  h(s) * \, \frac 1 {s^2+4}$$
$$X(s) =3\frac {s}{s^2+4}-\frac 12\frac {2}{s^2+4}+  \frac 12h(s) * \, \frac 2 {s^2+4}$$
Because $\mathcal{L^{-1}}(h(s))=\mathcal{L^{-1}}\mathcal{L}(f(t))=f(t)$ Using the convolution formula, we get that :
$$\boxed {x(t) =3\cos(2t)-\frac 12\sin(2t)+  \frac 12\int_0^t f(\tau) \, \sin(2(t-\tau))d\tau}$$
A: I think I found the solution already:
$$X(s)= \frac{\mathcal{L(f(t))}}{s^2+4} + \frac{3s-1}{s^2+4}$$
$$= \frac{\mathcal{L(f(t))}}{s^2+4} + \frac{3s}{s^2+4} + \frac{-1}{s^2+4}$$
Since the 'standard results of: $\frac{s}{s^2+4} = A\cos2t$ and $\frac{1}{s^2+4}= B\sin2t$ The first term can be arranged by the convolution theorem:
$$\frac{\mathcal{L(f(t))}}{s^2+4} = \frac{1}{2}\int_{0}^{t}\sin2(t-\tau)\cdot f(\tau)dt $$
combining this all:
$$X(t)= A \cos2t+B \sin2t +\frac{1}{2}\int_{0}^{t}f(\tau)\sin2(t-\tau)d\tau$$
