Non-linear differential equation

  • Thread starter pd1zz13
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  • #1
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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?
 

Answers and Replies

  • #2
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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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