Extremals

**bobby** · February 9th 2009, 05:53 AM

Calculate and sketch the extremals of integral (from 1 to b) of (x^2*y'^2 + 12y^2) dx which pass through (1,1) and where b>1.

There is a hint, to solve the Euler-Lagrange equation, try solutions of the form y = x^p but I don't see how that helps.

Any help would be appreciated.

Thanks.

**HallsofIvy** · February 9th 2009, 11:40 AM

Well, what is the Euler-Lagrange equation for this problem?

**bobby** · February 9th 2009, 12:05 PM

(x^2)y'' + 2xy' -12y = 0

???

**HallsofIvy** · February 11th 2009, 09:58 AM

Originally Posted by bobby

(x^2)y'' + 2xy' -12y = 0

???

Okay, and the other hint was to try $y= x^p$ . Then $y'= px^{p-1}$ and $y"= p(p-1)x^{p-2}$ . Putting those into the equation we have $x^2(p(p-1)x^{p-2}+ 2x(px^{p-1})- 12x^p= [p(p-1)+ 2p- 12]x^p= 0$ which will be true, for all x, if and only if $p(p-1)+ 2p- 12= p^2+ p- 12= (p-3)(p+4)= 0$ so the general solution to the Euler-LaGrange equation is $y(x)= Cx^3+ Dx^{-4}$ . Now, what does the Euler-LaGrange equation tell you about the extremals?

**bobby** · February 11th 2009, 01:57 PM

Thanks, so does that mean the coefficient of the first derivative is

Cx^3+ Dx^{-4}

? Not sure on this.

just wondering, did I get my ELeqn correct or is it meant to be 6y - xy' = 0 ? As I'm not sure if you're meant to differentiate both terms in d/dx ( 2x^2 * y') as y' is a function of x..?

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