Differentiating an "initial value problem"?

Question

I have the following ODE

$$y' = 2 - \sin(xy), \qquad 1 \le x \le 3$$

with initial condition $y(1)=-\frac 12$. I have to prove that $|y''(x)| \le 40$ over the domain $[1,3]$.

I thought perhaps just do partial differentiation on y' and then draw it on a graph using the x domain and show its always under 40, but that doesn't seem to work... any help? If it helps, the question is in the context of an Euler question. Cheers.

Answer 1

Following your idea of differentiating, note that

$$y'' = -\cos(xy) (xy' + y) \implies |y''| \le |x| |y'| + |y|.$$

Now $|x| \le 3$ and $|y'| \le 3$ as well (why?), so it remains to check how large $y$ can get. Since $y(1) = -1/2$ and $|y'| \le 3$, one can quickly show that $|y| \le 6.5$ for $x$ in this range.

Put it all together and you get an upper bound which is quite a bit better than $40$.

Answer 2

Differentiating the expression yields

$$y'' = -\cos(xy)\left[y+xy'\right]= -\cos(xy)\left[y+x(2-\sin(xy))\right].$$

Using the modulus and the triangular inequality:

$$\implies |y''| \leq |y+x(2-\sin(xy))| \leq |y|+|x||2-\sin(xy)|\leq |y|+3|x|\leq |y|+9.$$

The last thing that remains is finding a bound on $|y|$ in order to do this we notice that $y'>0$. Hence, the values of $y$ are rising all from $x=1$ to $x=3$. The maximal growth rate is $3$ which implies $y'=2-\sin(xy)\leq 3 \implies y(x)\leq 3(x-1)-1/2 \implies |y|\leq 5.5$.