non-linear differential equation


by pd1zz13
Tags: differential, equation, nonlinear
pd1zz13
pd1zz13 is offline
#1
Nov13-11, 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?
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jackmell
jackmell is offline
#2
Nov14-11, 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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