Solving an Euler Equation (Calculus of variations)...

I need to solve the Euler equation to make the following integral stationary:

$\displaystyle \int ( y'^2+\sqrt{y}) dx$

so I happily assume $\displaystyle F = y'^2+\sqrt{y} $ and find the partial derivative F wrt y' and F wrt y, put both into the Euler equation:

$\displaystyle \frac{d}{dx}(2y') -\frac{1}{2\sqrt{y}} = 0$

Route 1: simple rearrangement yield:

$\displaystyle y''- \frac{1}{4}y^{-\frac{1}{2}} = 0$

Route 2: intergate both side by x

$\displaystyle \int \sqrt{y} dy' = \frac{x}{4}$

Then I'm stuck in both.... there is probably a very simple way of getting the solution out, but my brain just can't turn the concer~ or I've already made a mistake early in my calculations?

Thanks for all the help!

Re: Solving an Euler Equation (Calculus of variations)...

Quote:

Originally Posted by

**QDsolarX** I need to solve the Euler equation to make the following integral stationary:

$\displaystyle \int ( y'^2+\sqrt{y}) dx$

so I happily assume $\displaystyle F = y'^2+\sqrt{y} $ and find the partial derivative F wrt y' and F wrt y, put both into the Euler equation:

$\displaystyle \frac{d}{dx}(2y') -\frac{1}{2\sqrt{y}} = 0$

Route 1: simple rearrangement yield:

$\displaystyle y''- \frac{1}{4}y^{-\frac{1}{2}} = 0$

Route 2: intergate both side by x

$\displaystyle \int \sqrt{y} dy' = \frac{x}{4}$

This does not strike me as valid. You were good up to the point where you wrote down your second-order DE. I would go this route: whenever you have a DE of the form

$\displaystyle y''=f(y),$

you can multiply through by $\displaystyle y'$ and integrate at least once immediately. In your case, you get

$\displaystyle y''y'=\frac{1}{4}\,y^{-1/2}y'$

$\displaystyle \frac{(y')^{2}}{2}=\frac{\sqrt{y}}{2}+C_{1}.$

Now you have a first-order DE. Can you continue from here?

Quote:

Then I'm stuck in both.... there is probably a very simple way of getting the solution out, but my brain just can't turn the concer~ or I've already made a mistake early in my calculations?

Thanks for all the help!