Non-linear differential equation

[pd1zz13]

The original equation is:

V'' - k*V^-1/2 = 0

I then said u = V' therefore u' = u(du/dV)

so the new equation is :

u(du/dV) = V^-1/2*k

Im a little rusty on separation of variables but I got a u in the final answer which means I have to integrate again since I need the final answer in terms of V.

The goal is to prove k is proportional to V^3/2

Am i doing this right?

 

[jackmell]

That's a common way to do it and you're left with:

u^2=4k\sqrt{v}+c

or:

\left(\frac{dv}{dt}\right)^2=4k\sqrt{v}+c

taking square roots, separate variable again and get:

\frac{dv}{\sqrt{4k\sqrt{v}+c}}=\pm dt

That however looks kinda' messy to integrate although Mathematica gives a nice expression for the left side but then you get an implicit expression for v in terms of t:

g(v)=\pm(t+c_2)

which is not solvable explicitly for v(t). Doesn't look like it anyway.