- #1

skrat

- 748

- 8

## Homework Statement

We have functional ##I(y)=\int_{0}^{2}{y}'(2+e^x{y}')dx## where ##y\in C^1(\mathbb{R})## and ##y(0)=0##. Calculate the extreme value.

## Homework Equations

## The Attempt at a Solution

I am having some troubles here... :/

From Euler-Lagrange equation we get ##\frac{\partial L}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial L}{\partial {y}'}=0## where first part is obviously ##\frac{\partial L}{\partial y}=0## and the second part tells me that ##\frac{\partial L}{\partial {y}'}=konst.=C##.

So ##(2+e^x{y}')+{y}'e^x=C##

##e^x{y}'=\frac{C}{2}-1=D##, where D is just another constant...

##{y}'=De^{-x}## so

##y=-De^{-x}+E##

Now we know that ##y(0)=0## so ##y(0)=-D+E=0## therefore ##E=D##. The second condition says that ##\frac{\partial }{\partial {y}'}L(2)=0##.

Than ##(2+e^x{y}')+{y}'e^x=0## so ##{y}'=0##.

From ##y=-De^{-x}+E## we get that ##{y}'=De^{-x}##, therefore ##{y}'(2)=De^{-2}=0## and this gives me ##D=0## (remember that ##E=D##).

So everything together tells that ##y## is actually ##y\equiv 0## and also ##I(y)=0##.I seriously doubt that this is right? :/