Given a first order differential equation: $ y' = y^2 +x $
Initial condition: $ y(0)= 0 $
Prove that the differential equation above has no solution on $[0,3]$.
I don't really understand how such equation has no solution on $[0,3]$.
Both $f(x,y) = y^2+x $ and $\frac{\partial f}{\partial y} $ are continuous at $[0,0]$ so the existence and uniqueness theorem says it has unique solution ?
I even used Matlab to graph the solution (click for full size image):
The red curve is the solution that satisfies $y(0)=0$.