calculus - Laplace Transform

Question: Use Laplace Transform to solve the following differential equation

$\ y''+y =sin(t); y(0)=1, y'(0)=-1 $

My try,where F(s) is the transform of f(t)=y(t)

$F(s)= \frac{1}{(s^2+1)^2} + \frac{s}{s^2+1} -\frac{1}{s^2+1} $

Then,using convolution for Laplace Transform in the first fraction and applying the transform in others

$y(t)= \int_0^t \sin(t-u)\sin(u)du + \cos t - \sin t $

My problem is here,because I'm reaching to a result that is different from the answer. My approach was

$$ \int_0^t \sin(t-u)\sin(u)du = \int_0^t (\sin(t)\cos(u)-\sin(u)\cos(t))\sin(u)du =$$

$$= \sin(t)\int_0^t\sin(u)\cos(u)du - \int_0^t\sin^2(u)\cos(t)du
=\frac{\sin^3(t)}{2}-\frac{\cos(t)}{2}\left(t-\frac{\sin(2t)}{2}\right) $$

And there is some problem in that resolution that I can't found. Someone can help please?

The answer is:

$ y(t)=\cos(t)-\frac{1}{2}\sin(t)-\frac{1}{2}t\cos(t) $

Answer

I really have no idea what you did there and why did you go for convolutions as the initial conditions are for zero and it is straightforward then, but the following is what I did:

Putting $\;\mathcal L\{y(t)\}(s)=Y(s)\;$ and applying the transform to the equation with the conditions, we get:

$$s^2Y(s)-s+1+Y(s)=\frac1{s^2+1}\implies (s^2+1)Y(S)=\frac1{s^2+1}+s-1\implies$$

$$Y(s)=\frac1{(s^2+1)^2}+\frac s{s^2+1}-\frac1{s^2+1}=\frac12\frac{2\cdot1^3}{(s^2+1)^2}+\frac s{s^2+1}-\frac1{s^2+1}\implies$$

$$y(t)=\frac12\left(\sin t-t\cos t\right)+\cos t-\sin t=\cos t-\frac12\sin t-\frac12t\cos t$$