TITLE: Is solution of $\frac{dy}{dx}=x^2+y^2 +1$ an odd function
QUESTION [6 upvotes]: I had this question in my exam which is of multiple select question (MSQ) type. I am sure about three options but confused about one.
Question is "Let $(-c,c)$ be the largest open interval in $R$ ($c$ >0) on which the solution $y(x)$ of the differential equation $\frac{dy}{dx}=x^2+y^2 +1$ with initial condition $y(0)=0$ exists and is unique. Then which of the is/are true?"
Options were
(A) $y(x)$ is an odd function on $(-c,c)$
(B) $y(x)$ is an even function on $(-c,c)$
(C) $(y(x))^2$ has a local minimum at $0$
(D) $(y(x))^2$ has a local maximum at $0$
Clearly $\frac{dy}{dx}>0$, so $y$ is strictly increasing. Given that $y(0)=0$, we then have $y(x)<0$ for $x<0$ and $y(x)>0$ for $x>0$. Using this, options (B) and (D) get discarded and option (C) looks correct. But not getting any idea about option (A). Also I tried to solve this ODE but no method is working that I know.
Any hint or help. Thanks.

REPLY [3 votes]: (C) is also correct. By implicit differentiation, $\frac12(y^2)' = yy' = y(x^2+y^2+1)$. So $\frac12(y^2)'' = y'(x^2+y^2+1) + y(2x+2yy')$. From this we deduce that $(y^2)' = 0$ at $x=0$ and $(y^2)'' = 1$ at $x=0$, and so $y^2$ has a local minimum at $0$.