alejandrito20 Posted February 11, 2010 Share Posted February 11, 2010 [math](\frac{du}{dy})^2=A+Be^{2u}+C \sqrt{D+Ee^{4u}}[/math] where A,B,C,D,E are nonzero Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted February 11, 2010 Share Posted February 11, 2010 This shouldn't be too hard. Show us how you'd start it. You'll have to use the chain rule a few times. Link to comment Share on other sites More sharing options...
alejandrito20 Posted February 11, 2010 Author Share Posted February 11, 2010 (edited) 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 Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted February 11, 2010 Share Posted February 11, 2010 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. Link to comment Share on other sites More sharing options...
Dave Posted February 12, 2010 Share Posted February 12, 2010 Jeez, that looks like a pretty nasty equation. I doubt you're easily going to find any solutions via standard methods. Link to comment Share on other sites More sharing options...
Amr Morsi Posted March 28, 2010 Share Posted March 28, 2010 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. Link to comment Share on other sites More sharing options...
the tree Posted March 30, 2010 Share Posted March 30, 2010 (edited) 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. Link to comment Share on other sites More sharing options...
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