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

Attempt:=

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

Not sure what to do next.

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2 Answers 2

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

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  • $\begingroup$ How did you get $x-\frac{3}{2}$ and $3x-\frac{7}{2}$? $\endgroup$
    – user572780
    Feb 25, 2019 at 19:49
  • $\begingroup$ 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. $\endgroup$
    – Delta-u
    Feb 25, 2019 at 19:53
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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.

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

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