On the open interval $(-c,c)$, where $c$ is a positive real number, $y(x)$ is an infinitely differentiable solution of the differential equation $$\frac{dy}{dx}=y^2-1+\cos x$$ with the initial condition $y(0)=0$. Then which one of the following is correct?
(A) $y(x)$ has a local maximum at the origin
(B) $y(x)$ has a local minimum at the origin
(C) $y(x)$ is strictly increasing on the open interval $(-\delta,\delta)$ for some positive real number $\delta$
(D)$y(x)$ is strictly decreasing on the open interval $(-\delta,\delta)$ for some positive real number $\delta$
My Attempt
Since $y(0)=0$ and function is differentiable and hence continuous so it should be very close to $0$ in the neighborhood of $x=0$. So $y^2+\cos x$ could be greater than $1$ in the neighborhood of $x=0$.On the other hand the opposite may as well be true.So, $(C)$ or $(D)$ may be correct. But what will be actual/logical explanation.
The equation and initial condition are invariant under inflection $(x,y)\mapsto(-x,-y)$. Thus this specific IVP solution has to be an odd-symmetric function.
The leading lowest-degree terms in $y=a_mx^m+...$ has to satisfy $$ ma_mx^{m-1}+...=a_m^2x^{2m}+...-\frac12x^2+... $$ which implies $m=3$, $3a_m=-\frac12$.