I've got the equation which I am supposed to use laplace transforms to solve:
$$ y'' + y = e^t + tsin(t), ~~~~~y(0) = 3/2,~~~~~y'(0) = 5/2 $$
Using laplace transforms, I come up with the equation:
$$ Y(s) = {1 \over (s-1)(s^2+1)} + {2s \over (s^2+1)^3} + {3s+5 \over 2(s^2+1)}$$
I can use partial fraction and inverse laplace to solve most if it:
$$ y(t) = \frac{1}{2}e^t +cos(t) + 2sin(t) + L^{-1} \left[\frac{2s}{(s^2+1)^3} \right]$$
I've been trying to use convolution:
$$ L^{-1} \left[ \frac{s}{(s^2+1)^2} \right] = \frac{t}{2}sin(t) ~~~~~L^{-1} \left[ \frac{2}{(s^2+1)} \right] = 2sin(t) $$
$$ \left(2sin(t) * \frac{t}{2} sin(t)\right) = \int_0^t \frac{\tau}{2}sin(\tau) \cdot 2sin(t-\tau) d\tau$$
I just can't seem to get the two terms I'm missing. Is there a better way to approach the laplace or convolution? Or do I just need to make sure I'm doing the integral correctly?
Symbolab answer to check for correctness:
$$y=-\frac{1}{4}t^2\cos \left(t\right)+2\sin \left(t\right)+\frac{e^t}{2}+\frac{1}{4}t\sin \left(t\right)+\cos \left(t\right)$$
Best Answer
What if we write:
$$q(s) = \dfrac{2s}{(s^2+1)^3} = -\dfrac{1}{2} \dfrac{d}{ds} \dfrac{1}{(s^2+1)^2}$$
Now, we ca write:
$$\mathcal{L}^{-1} (q(s)) = \mathcal{L}^{-1} \left(-\dfrac{1}{2} \dfrac{d}{ds} \dfrac{1}{(s^2+1)^2} \right)$$
Let's do it with a Laplace Table first, items $11$ and $30$:
$$ \mathcal{L}^{-1} \left(-\dfrac{1}{2} \dfrac{d}{ds} \dfrac{1}{(s^2+1)^2} \right) = \dfrac{1}{4} t\left(\sin t - t \cos t\right)$$
We can also use the convolution method that you tried.
$$ \mathcal{L}^{-1} \{ F(s) G(s) \} = (f * g)(t)$$
We set (try other variants):
$F(s) = \dfrac{2}{(s^2+1)^2} \implies \mathcal{L}^{-1}(F(s)) = f(t) = \sin t -t \cos t \\ G(s) = \dfrac{s}{s^2+1} \implies \mathcal{L}^{-1}(G(s)) = g(t) = \cos t$
We now have:
$$\displaystyle \int_0^t f(t-v) g(v)~dv = \int_0^t f(v) g(t-v)~dv = \dfrac{1}{4} t( \sin t- t \cos t)$$
Looks like you know how to do the rest of the problem.
Update
For the integral, we have $f(v) g(t-v) = \cos (t-v) (\sin v -v \cos v)$. When we integrate, we get:
$$\displaystyle \int \cos (t-v) (\sin v -v \cos v)~dv = \dfrac{1}{8} \left(-2 v^2 \cos t+2 v (\sin (t-2 v)+2 \sin t)-3 \cos (t-2 v)\right)$$
Evaluating this at $v = 0$ and accounting for subtraction, we get:
$$\dfrac{3}{8} \cos t$$
Evaluating this at $v = t$ and accounting for subtraction, we get:
$$\dfrac{1}{8} \left(-2 t^2 \cos t+2 t \sin t-3 \cos t\right) $$
Adding those two, gives the result above.