Is $y^2$ an even function?

Question

yesterday I asked this question

the solution of this problem says that since $y^2$ is an even function and so is $cosx-1$ therefore $y'$ is an even function which implies $y$ is odd function and from there on the solution builds up

I can not understand how is $y(x)^2$ even??

here is the full solution https://www.youtube.com/watch?v=WI3UBYJ21iw&t=98s

and people here on stackexchange told he that the solution is wrong, which is what I was thinking too but today I encounter the same approach again and both the times the answer though this approach is correct which made me confused, how is $y^2$ even?

I am getting more confused now here is one more example where $y^2$ is taken to be even

this is the solution https://www.youtube.com/watch?v=vnqOCpqUebQ

what's going on???

I am also looking for the correct solution of the second problem, the first problem I was able to solve myself

Answer 1

As far as I know, there is no way to tell that $y^2$ is an even function a priori. Perhaps whoever presented solution confused polynomials in $x$ with even powers with this case and jumped to conclusion that $y^2$ is always even. This is obviously false for a function like $y(x) = x^2 + x$.

A way to solve the first problem that I can see immediately is to use higher derivative test for extrema. We have

$$y' = y^2 - 1 + \cos x \implies y'(0) = y(0)^2-1+\cos 0\stackrel{y(0)=0}{=} 0$$ $$y''= 2yy'-\sin x \implies y''(0) = 2y(0)y'(0)-\sin 0 = 0 $$ $$y''' = 2(y')^2 + 2yy'' - \cos x \implies y'''(0) = 2y'(0)^2+2y(0)y''(0) - \cos 0 = -1$$

We conclude that $y$ has inflection point at the origin, so A and B are false. Furthermore, since $y$ is infinitely times differentiable, in particular $y'''$ is continuous, so there exists $\delta > 0$ such that $y''' < 0$ on $(-\delta,\delta)$. Therefore, $y''$ is strictly decreasing on $(-\delta,\delta)$, so $y'' > 0$ on $(-\delta,0)$ and $y''<0$ on $(0,\delta)$. That means that $y'$ is strictly increasing on $(-\delta,0)$ and strictly decreasing on $(0,\delta)$. Thus, $y' < 0$ on $(-\delta,\delta)\setminus\{0\}$. Therefore, the correct answer is D.

Visualizing this argument:

\begin{array}{c| l c c c r} & -\delta && 0 && \delta\\ \hline y''' && - && -\\ \hline y'' && \searrow && \searrow \\ && + && - \\ \hline y' && \nearrow && \searrow \\ && - && - \\ \hline y && \searrow && \searrow \end{array}

Answer 2

Consider the symmetric of the solution wrt the origin, i.e. $z(x)=-y(-x)$. You have $z'(x)=y'(-x)$ and the differential equation

$$\frac{\mathrm dz}{\mathrm dx}=y'(-x)=y^2(-x)-1+\cos(-x)=z^2(x)-1+\cos(x)$$

Therefore $z$ satisfies the same DE as $y$. The same initial condition $y(0)=z(0)=0$ holds, therefore the solutions are identical, and $y(x)=-y(-x)$, so $y$ is odd.

A maybe more intuitive way to view this: the slope is a function of $y$ and $x$ on a $y$-$x$ plot. And the slope at $(x,y)$ is the same as the slope at $(-x,-y)$ (because $y^2$ is an even function... of $y$). By symmetry of the slope field wrt the origin, the solution has to be symmetric too, as it passes through the origin.

With the same approach you should be able to solve the other problem.