This is @lacypennelll problem. Kindly somebody help :)

Question

This is @lacypennelll  problem. Kindly somebody help :)

ganeshie8 · Answer

Are you saying y' is undefined at x=0, y=3 ?
@lacypennelll

ganeshie8 · Answer

I don't think that is an issue as x=0 gives you $y' = 0$

ganeshie8 · Answer

These domain restrictions of DE solutions are a bit tricky. Here is what I think :

The general solution is
$$y = \sin(c+x^2)$$
Clearly $y$ values are restricted to $[-1,1]$.

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=solve+y%27+%3D+2x+sqrt%281-y^2%29

ganeshie8 · Answer

I think so, but I could be terribly wrong here... 

@Loser66

ganeshie8 · Answer

go through when free http://ecademy.agnesscott.edu/~lriddle/apcalculus/apcentral-domain.pdf

lacypennelll · Answer

That makes sense! Thank you so much!

ganeshie8 · Answer

Np, please wait and take @Loser66 's opinion too

across · Answer

You guys are right. This is a nice application of the existence theorem. $y'=F(x,y)$ is continuous on $\mathbb R\times[-1,1]$, which does not include $(0,3)$. ;-P

loser66 · Answer

To me, use separable to solve the ODE, you get $sin^{-1} y = x^2 +C$ that gives you
$y = sin(x^2+C) $, hence, with $y(0) = 3$ sin c =3 , that gives you no c satisfies it. :)

ganeshie8 · Answer

Yaaay! thank you both :D

loser66 · Answer

hihihi...