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

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

Attempt:=

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

Not sure what to do next.

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

• How did you get $x-\frac{3}{2}$ and $3x-\frac{7}{2}$? Feb 25, 2019 at 19:49
• You have for all $s \in [1,3]$, $1 \leq y'(s) \leq 3$. So integrating for $s$ between $1$ and $x$ you obtain $$\int_1^x 1 ds \leq \int_1^x y'(s) ds \leq \int_1^x 3 ds$$ i.e $$x-1 \leq y(x)-( -\frac{1}{2}) \leq 3x -3$$ which leads to the desired inequality. Feb 25, 2019 at 19:53

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.

• Is $|y''(x)| = |-\cos(xy)(y + xy')|$ correct? Feb 25, 2019 at 19:08
• @user572780 Yes, I apologize, had a typo. Feb 25, 2019 at 19:09