Euler's Equations for Extremas of J: y=C*e^x

[lylos]

Homework Statement

For the functional $$J(y(x))=\int^{x1}_{x2}F(x,y,y')dx$$, write out the curve $$y=y(x)$$ for finding the extremas of J where $$F(x,y,y')=y'^2+y^2$$.

Homework Equations

Euler's Equations:
$$\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'}=0$$
$$\frac{\partial f}{\partial x} - \frac{d}{dx}(f-y' \frac{\partial f}{\partial y'})=0$$

The Attempt at a Solution

Using $$\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'}=0$$,
$$\frac{\partial f}{\partial y}=2y$$
$$2y=\frac{d}{dx}\frac{\partial f}{\partial y'}$$
$$2y=\frac{d}{dx}2y'$$
$$y=\frac{d^2y}{dx^2}$$
$$y=C*e^x$$ Where C is a constant.

Is this correct? Using the 2nd equation, I get an ugly answer that involves Sinh.

 

[zaglamir]

Yo,

The general solution will be y=Ce^(lambda)x.

So you do y''-y=0

Which will look like: c(lambda)^2*e^x-c*e^x=0.

You can find lambda to be + or - 1. So your general solution will be C1*e^x+C2*e^-x=y.



Also, on the rest of our Homework III and IV, you can assume that it is a constant even if it has an x in it, Sophie said so.

 

[thomate1]

Your solution is correct. However, you can simplify it further by using the second equation to eliminate the second derivative of y.

Starting with \frac{\partial f}{\partial x} - \frac{d}{dx}(f-y' \frac{\partial f}{\partial y'})=0, we can substitute in the given values for F(x,y,y') and solve for y.

\frac{\partial f}{\partial x} - \frac{d}{dx}(y'^2+y^2-y'2y)=0
\frac{\partial f}{\partial x} - \frac{d}{dx}(y'^2-y^2)=0
2y'y''-2yy'=0
y''-y=0
y=C_1e^x+C_2e^{-x}

This is equivalent to your solution, but without the need for the second derivative.