# non-linear differential equation

by pd1zz13
Tags: differential, equation, nonlinear
 P: 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?
 P: 1,653 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.

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