# Calulus of variations Euler Lagrange equation

How can I find an extremal for the following functional: $$J[y] = \int_0^1 e^y (y')^2 dx$$

with the boundary conditions $y(0)=1, y(1)=\log(4)$? I have tried computing the Euler-Lagrange equation but it becomes overly complicated.

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The E-L equation is $- e^y (y')^2 - 2 e^y y'' = 0$. Divide by $e^y$ and first solve the first-order equation for $y'$, then integrate to get $y$. Then put in the boundary conditions.
Hint: The Lagrangian $$L(y,\dot{y})~=~ e^{y} \dot{y}^2$$ does not depend explicitly on the independent variable. Noether's theorem then states that the energy function $$h(y,\dot{y})~=~\dot{y} \frac{\partial L(y,\dot{y})}{\partial \dot{y}}- L(y,\dot{y})~=~\ldots ~=~L(y,\dot{y})$$ is conserved, and hence provides a first integral $$e^{y/2} \dot{y}~=~{\rm const}.$$ (This fact is also known in mathematics as the Beltrami identity.)
The Euler-Lagrange equation simplifies down to $y''+(y')^{2}=0$. Set $u=y'$ to get a seperable ODE, then integrate to find $y$.