I am solving a problem on calculus of variation in which $F(x,y,y')$ is given as $F(x,y,y')=e^yy'^2$
After solving Euler equation I got this $2y'' +2y'-y'^2=0$.
I don't know how to proceed further. Please guide me. Thanks in advance.
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1you might start by trying the substitution $u=y'$ so that you have a first-order differential equation? – David Holden Dec 05 '15 at 13:19
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@david holden I did that but the solution is getting complicated. – zafran Dec 05 '15 at 13:23
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To solve the following differential equation you can use reduction of order here by letting $u = y',\ u' = y''$: $$2y'' +2y'-(y')^2=0$$ $$2u' + 2u - u^2 = 0$$ $$2\frac{du}{dx} = u^2 - 2u$$ $$\int\frac{2}{u^2 - 2u}du = \int dx$$ $$\int\left(\frac{1}{u-2}-\frac{1}{u}\right)du = x + C$$ $$\ln|u-2| - \ln|u| = x+C$$ $$\ln\left|\frac{u-2}{u}\right| = x+C$$ $$\frac{u-2}{u} = Ce^x$$ $$u-2 = uCe^x$$ $$u(1-Ce^x) = 2$$ $$u = \frac{2}{1-Ce^x}$$ Since $u = y'$, we can substitute to get: $$y' = \frac{2}{1-Ce^x}$$ $$y = \int\frac{2}{1-Ce^x}$$ Can you take it from here?
Varun Iyer
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Thank you Varun. I got the same but after integrating R.H.S I am not getting desired answer. – zafran Dec 05 '15 at 14:21
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Thank you. Actually it was an incomplete question but it doesn't matter because I am having problem with solving this differential equation but it is not matching with given answer. – zafran Dec 05 '15 at 14:53