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[alejandrito20]

[math](\frac{du}{dy})^2=A+Be^{2u}+C \sqrt{D+Ee^{4u}}[/math]

 

where A,B,C,D,E are nonzero

[Cap'n Refsmmat]

This shouldn't be too hard. Show us how you'd start it. You'll have to use the chain rule a few times.

[alejandrito20]

i have tried to do

 

[math]\int\frac{du}{\sqrt{A+Be^{2u}+C\sqrt{D+Ee^{4u}}}}=\int dy[/math]

 

but, this integral ........????????

 

chain rule is [math]2\frac{du}{dy}\frac{d^2u}{d^2y}=2Be^{2u}\frac{du}{dy}+2CE\frac{du}{dy}\frac{e^{4u}}{\sqrt{D+Ee^{4u}}}[/math]????????

Edited February 11, 2010 by alejandrito20

[Cap'n Refsmmat]

Oh, sorry, I got totally confused there. Ignore me.

 

I don't see how this could be easily integrated. You'll have to try another method besides separation of variables.

[Dave]

Jeez, that looks like a pretty nasty equation. I doubt you're easily going to find any solutions via standard methods.

[Amr Morsi]

Do you have a relation D>>E or E>>D? This can be used in an approximation which makes it possible to integrate to a closed form.

[the tree]

That's a good point. For really. really high [imath]u[/imath] it would approximate to:

[math]\left( \frac{\mbox{d}u}{\mbox{d}x} \right)^2 = B e^{2 u} + C \sqrt{E e^{4 u}} [/math]

[math]\left( \frac{\mbox{d}u}{\mbox{d}x} \right)^2 = (B + C \sqrt{E}) e^{2 u} [/math]

[math] \frac{\mbox{d}u}{\mbox{d}x} = \left( \sqrt{B + C \sqrt{E}} \right) e^{u} [/math]

[math]e^{u} = -x\sqrt{B + C \sqrt{E}}+F[/math]

Edited March 30, 2010 by the tree
Consecutive posts merged.