Difficulty in solving calculus of variations problem.

Question

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.

Answer 1

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?