Proving that $y''(x)=\cos(x)+y^3(x)+3x\,y^2(x)\,y'(x)>0.$ Let $y'(x)=\sin(x)+x\,y^3(x)$ and $y(0)=0$.
How can I prove that $y''(x)>0$ for small $x$?
Got the information $v'=\sin(x), v(0)=0$ and $w'=x\,y^3(x), w\left(\dfrac{\pi}{2}\right)=\dfrac{\pi}{2}.$
My way was:
$$y''(x)=\cos(x)+y^3(x)+3x\,y^2(x)\,y'(x).$$
It's $\mathrm{cos}(x)>0$ for $x \in \left[0,\dfrac{\pi}{2}\right]$.
Now I don't know how to show that it's still positive for $y''(x)$ since I don't know what $y(x)$ is.
How can it be shown?
 A: Intuitively, it makes sense: note that $y'(0)=0,$ and as $x$ gradually increases from $0,$ everything on the RHS of $y'(x)=\sin(x)+x\,y^3(x)$ is increasing at least up until $\pi/2.$ So you would certainly expect $y''(x)$ to be positive at least on $(0,\pi/2).$
Note that $y''(0)=1.$ If we can argue that $y''$ is continuous, we are done, since then $y''$ wouldn't be able to "jump down" below zero instantaneously. To put it rigorously: $\forall\,\varepsilon>0$ there exists $\delta>0$ such that if $0\le x<\delta,$ then $|y(x)-1|<\varepsilon.$ So, just take any $\varepsilon<1.$
But now notice that we can plug in for $y'$ into your expression for $y'':$
\begin{align*}
y''(x)
&=\cos(x)+y^3(x)+3x\,y^2(x)\,y'(x)\\
&=\cos(x)+y^3(x)+3x\,y^2(x)\left[\sin(x)+x\,y^3(x)\right].
\end{align*}
This shows us that if $y$ is continuous, $y''$ is continuous.
If we write $y'=f(y,x),$ then $f$ is continuous on any rectangle, and we can invoke the Cauchy-Peano Existence theorem (Theorem 1.2 in Coddington and Levinson) to show that there exists a $C^1$ solution satisfying the initial condition. Theorem 2.3 in the same book can be used to show uniqueness, since, while $f$ is not Lipschitz on $\mathbb{R},$ it is Lipschitz in $y$ on a small interval containing $0.$
This proves the result.
