# A simple integration problem

I'm stuck with a simple integration problem. Haven't done this in a while so I've gone a little rusty. Little help would be appreciated. Well anyway, here goes:

$dy/dt=\sqrt{{u}{y}+{v^2}}$

My attempt:

for the sake of simplicity let $dy/ty = T$

$T^2=uy+v^2$

$T^2dy/dt=(uy+v^2dy)/dt$

$T^2dy=uy+v^2dy$

$T^3/3+C=u^2/2*y^2/2+v^3/3$

$T^3/3=u^2/2*y^2/2+v3/3-C$

And I'm stuck. Quite frankly I'm not even sure that the two last steps were right.

-
What are $\;u,v,k\;$ and what their relation with $\;t\;$ ?? – DonAntonio Mar 27 '14 at 11:30
$\int uy=(\int u\,du)(\int y\,dy)$ is certainly not correct. – Gerry Myerson Mar 27 '14 at 11:38
The calculations don't make much sense I'm afraid. Assuming $u$ and $v$ are constants, you could write it as : $$\int\dfrac{dy}{\sqrt{uy + v^2}} = \int dt = t$$ and integrate from here. – user88595 Mar 27 '14 at 11:57

As suggested by user88595, the differential equation is separable since you can write (as user88595 wrote) $$\int\dfrac{dy}{\sqrt{uy + v^2}} = \int dt = t$$ in which you can recognize a quite classical integral (if you do not, use a change of variable such that $uy=v^2z$. Doing so, you should arrive to $$\frac{2 \sqrt{u y+v^2}}{u}=t+C$$ which, in turn, leads to $$y=\frac{t^2 u^2+2 C t u^2+C^2 u^2-4 v^2}{4 u}$$