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Prove that all maximal solutions are defined in $\mathbb{R}$ for $y'=\cos(x+y)$. Find maxima, minima and critical points.

I managed to find a solution: Let $u=x+y$, then $\displaystyle\frac{du}{dx}=1+\displaystyle\frac{dx}{dy}\implies\displaystyle\frac{du}{dx}=1+\cos u$. Separating variable I get $\displaystyle\frac{1}{1+\cos u}du=dx$ then $\displaystyle\int (1+\cos u)^{-1}du=\displaystyle\int dx \implies x=\cot(u/2)+C$.

Using $u=x+y$ is $x-C=\cot[(x+y)/2]\implies\displaystyle\frac{1}{x-C}=\tan[(x+y)/2]\implies \displaystyle\frac{x+y}{2}=\arctan\left(\displaystyle\frac{1}{x-C}\right)$.

Then the solution is given by $y=2\arctan\left(\displaystyle\frac{1}{x-C}\right)-x$

To find critical points I calculate $y'$ for the last form: $y'=\displaystyle\frac{1}{1-(x+C)^2}-1$, so the critical points appear when $x=C$.

For maxima and minima I calculated $y''=\displaystyle\frac{x+C}{1-(x+C)^2}$ but this is zero again.

Also, can someone explain the first question given?. I don't understand what it means "prove that all maxima solution are defined"; does it means that for every $C$ exists a maxima?

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  • $\begingroup$ I think you made a mistake in your integration: $\int(1+\cos u)^{-1} du = \tan(u/2) + C$ $\endgroup$ – amcalde Sep 26 '14 at 14:56
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See my comment,

I get that $$y = 2\arctan(x-C) -x,\qquad y' = \frac{2}{1+(x-C)^2} - 1$$

Maxima and minima come from $0 = y'$ and this happens when $ x = C \pm 1 $. Plugging these back into $y(x)$ we get: $$y(C\pm1) = 2\arctan(\pm1)-C\mp1=\pm\frac{\pi}{2}\mp1-C$$ So the maximum of $y$ occurs at $x=C+1$ and the minimum at $x=C-1$.

Also $y'' = 0$ exactly at $x = C$.

The graph of $y$ looks mostly like the graph of ArcTan except it is tilted downward at a 45 degree angle. The two bumps of ArcTan remain and give you two sections of the graph, one concave up and one concave down.

I think the question just means that for any $C$, you have a perfectly nice function. With a well defined local maximum at $\{C+1,\frac{\pi}{2}- 1 -C\}$ and a local minimum at $\{C-1,\frac{-\pi}{2} + 1 -C\}$, with the inflection point exactly at $x=C$.

Below I plotted $y$ for $C = \pm1,$ and $ C=-3$.

enter image description here

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