Prove $|y''(x)| \leq 40$ for all $x \in [1,3]$.

Question

$$y' = 2 - \sin(xy), \qquad\quad 1 \leq x \leq 3, \qquad\quad y(1) = -\frac{1}{2}$$

Attempt:=

$$|y''(x)| = |-\cos(xy)(y + xy')| \leq |y + xy'|$$

Not sure what to do next.

Answer 1

As already mentioned, using the fact that $-1 \leq \sin \leq 1$ you have $$1 \leq y' \leq 3$$ so integrating between $1$ and $x$ we obtain $$x-\frac{3}{2} \leq y(x) \leq 3x-\frac{7}{2}$$ and in particular $$-\frac{1}{2} \leq y(x) \leq \frac{11}{2}.$$

And you can easily obtain $$| y +x y'| \leq \max(\frac{1}{2},\frac{11}{2}+3*3) = \frac{29}{2} <40$$

Answer 2

First of all, the IVP $$\begin{cases}y' = 2- \sin(xy), \quad x \in [1,3] \\ y(1) = -1/2 \end{cases}$$ does not have a closed form solution. This means that we should find another way around.

Note that $y'(x) = 2 - \sin(xy)$. But we know that $-1 \leq \sin(x) \leq 1 \implies -1 \leq \sin(xy) \leq 1$.

This means that $y'(x) > 0 \forall x \in [1,3]$. That means that $y(x)$ is strictly increasing.

Now, use a numerical method for the IVP to approximate $y(3)$ since you have $y(1)$ and use that to find a bound for $|y''(x)| = |-(xy'(x)+y(x))\cos(xy)|$.

Note : I added the usage of a Numerical Method since I saw your "numerical-methods" tag.