
help with a problem
Hey guys, i just started a new course in differential equations and im kinda having a hard time. Heres one of the problems I am having difficulty with:
Verify that the piecewisedefined function
$\displaystyle y=\left\{\begin{array}{cc}x^2,&\mbox{ if }
x < 0\\x^2, & \mbox{ if } x\geq 0\end{array}\right. $
is a solution of the differential equations $\displaystyle xy'2y = 0 $ on $\displaystyle (\infty , \infty)$
Thanks!

What have you tried so far?

Im not even sure how to start this problem. Is there a clue you can give me?

Well, you're not asked to solve the DE, but to verify the solution. So, I would plug the given solution into the DE and see if you get an equality. What do you get when you follow that path?

hmph. i just did:
if $\displaystyle x = 1 $
$\displaystyle 2x^2 +2x^2 = 0$
$\displaystyle 2 + 2 = 0 $
$\displaystyle 0 = 0 $
$\displaystyle if x = 1 $
$\displaystyle 2x^2 2x^2 = 0 $
$\displaystyle 2  2 = 0$
$\displaystyle 0=0 $
Is this what its asking for?

Not quite. You cannot just choose two values of x. If x is any positive number, then $\displaystyle y= x^2$ so that y'= 2x. The equation xy'2y = 0 becomes $\displaystyle x(2x) 2x^2= 2x^2 2x^2= 0$ so the equation is satisfied for any positive x. If x is negative, $\displaystyle y= x^2$. Do the same with that. Most importantly, what happens at x= 0? In particular what is the derivative of y at x= 0? (You can't just differentiate $\displaystyle x^2$ and set x= 0 because the derivative involves a limit from both sides: If x= 0 and h is positive, then x+h= 0+ h= h is positive, if h is negative, x+ h= h is negative.
$\displaystyle \lim_{h\to 0^+} \frac{y(h) y(0)}{h}= \lim_{h\to 0}\frac{2h^2}{h}= ?$
$\displaystyle \lim_{h\to 0^} \frac{y(h) y(0)}{h}= \lim_{h\to 0}\frac{2h^2}{h}= ?$
If those two limits are the same, that is the dervative at x= 0. If they are not, the function is not differentiable at x= 0.