Laplace Transform of derivative squared I'm trying to solve a problem while I'm studying Control Theory and I came up with a difficult question.
$
\mathcal{L}\left[y'(t)^2 \right]
$
Basically I need to find the Laplace Transform of this problem. In essence the differential equation I am attempting to solve looks like this,
$
y'(t) =a\,\sqrt{y(t)}
$
I couldn't find anything on regular Laplace Tables and I tried doing the integral on my own but it led me nowhere.
I could go both ways, either get the transform of $y'(t)^2$ or the transform of $\sqrt{y(t)}$.
 A: I can remove the derivative for you, but my expression still involves $ \mathcal L(y(t)^2) $.  Maybe you can work it as a convolution?
Anyways, check my work, but using integration by parts, with $ u = e^{-st} $ and $ v = (y'(t))^2 $, and the fact that $\int (y(t))^2 dt = y'(t)y(t) + \int y'(t)y(t)dt $, and $\int y'(t)y(t)dt = y(t)^2/2 $
$$\int _0^\infty e^{-st} (y'(t))^2 dt =  e^{-st}y(t)(y'(t)-y(t)/2)|^\infty_0 + s\int_0^\infty e^{-st}y(t)(y'(t)-y(t)/2) dt $$ 
Since
$$\int_0^\infty e^{-st}y(t)y'(t)dt = e^{-st}(y(t))^2/2|^\infty_0 + s\int_0^\infty e^{-st}(y(t))^2/2dt$$  
Then, under condition that the following limit applies for arbitrary constants A and B (they should for polynomial t-space)  $$\lim_{t\to \infty} e^{-st}(A*y(t)y'(t) + B*(y(t))^2) = 0$$
$$\int _0^\infty e^{-st} (y'(t))^2 dt =  y(0)y'(0) - \frac{(s+1)y(0)^2}{2} + \frac{s(s-1)}{2}(\int_0^\infty e^{-st}(y(t))^2 dt = \mathcal L(y(t)^2))  $$ 
A: Hint:
$$\frac{y'(t)}{\sqrt{y(t)}}=a \implies 2(\sqrt{y(t)})'=a$$
Solving for $y(t)$ will lead to a polynomial in $t$. Can you take it from there?
