# Non-linear differential equation

1. ### pd1zz13

3
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?

2. ### 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.