Calculating derivatives for the Euler equation

[mishima]

This is a calculus of variations problem from Boas chapter 9. I seem to be misunderstanding something with differentiation. Given

$$F=(1+yy')^2$$

then

$$\frac {\partial F} {\partial y'}=2(1+yy')y$$

and

$$\frac {\partial F} {\partial y}=2(1+yy')y' .$$

Now this one I am not so confident on...

$$\frac {d}{dx} \frac {\partial F} {\partial y'}=2y'+(4yy'^2+2y^2y'')$$

because after Euler's equation,

$$\frac {d}{dx} \frac {\partial F} {\partial y'} - \frac {\partial F} {\partial y}=0$$

I get this strange differential equation

$$2yy'+2y^2y''=0.$$

Am I making a mistake with the derivatives or just not realizing the correct technique for the final differential equation?

 

[BvU]

I see a mistake in the subtraction, not in the derivatives

 

[mishima]

Ah, thanks...forgot the exponent there.

$$2yy'^2+2y^2y''=0$$

Good to know the derivatives are correct. I guess this diff eq is a form I haven't encountered in the text, or is a familiar form in disguise. Ill concentrate in that direction.

 

[BvU]

Maybe it becomes less intimidating if you leave out a common factor $2y$

 

[mishima]

Got it, I had forgotten the square in my notes as well...thanks.

$$x=ay^2+b$$