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Nonlinear differential equation 
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#1
Nov1311, 06:11 PM

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? 


#2
Nov1411, 05:12 AM

P: 1,666

That's a common way to do it and you're left with:
[tex]u^2=4k\sqrt{v}+c[/tex] or: [tex]\left(\frac{dv}{dt}\right)^2=4k\sqrt{v}+c[/tex] taking square roots, separate variable again and get: [tex]\frac{dv}{\sqrt{4k\sqrt{v}+c}}=\pm dt[/tex] 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: [tex]g(v)=\pm(t+c_2)[/tex] which is not solvable explicitly for v(t). Doesn't look like it anyway. 


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