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I'm trying to solve this:

Find all general maximum solutions of this equation

[tex]

y'(2-e^{x}) = -3e^{x}\sin y\cos y

[/tex]

First, there are some singular solutions:

If

[tex]

y \equiv k\frac{\pi}{2}

[/tex]

Then right side is zeroed and so is the left.

To convert it to the separate form, I need to divide the equation with [itex]2-e^{x}[/itex]. But I can do this only if [itex]x \neq \log 2[/itex]

Otherwise I get

[tex]

y' = -\frac{3e^{x}}{2-e^{x}}\ \sin y\cos y

[/tex]

And leaving out the singular solutions, I can continue this way

[tex]

\frac{y'}{\sin y\cos y}} = -\frac{3e^{x}}{2-e^{x}}

[/tex]

[tex]

\int \frac{dy}{\sin y\cos y} = -3\int \frac{e^{x}}{2-e^{x}}\ dx

[/tex]

After evaluating the integrals, I got

[tex]

\log \left| \tan y \right| = \log \left|\left(2-e^{x}\right)^3\right| + C

[/tex]

Which I wrote in another way so that I could get rid of those logarithms

[tex]

\log \left| \tan y \right| = \log \left|\left(2-e^{x}\right)^3\right| + \log e^{C}

[/tex]

[tex]

\log \left| \tan y \right| = \log \left|e^{C}\left(2-e^{x}\right)^3\right|

[/tex]

And thus

[tex]

\tan y = e^{C}\left(2-e^{x}\right)^3

[/tex]

[tex]

y = \arctan \left[e^{C}\left(2-e^{x}\right)^3\right] + k\pi

[/tex]

Proof showed that this result is correct, but what remains to solve is the domain of [itex]x[/itex] for which it applies.

Even for [itex]x = \log 2[/itex] the equivalence holds, so I would say that

[tex]

y = \arctan \left[e^{C}\left(2-e^{x}\right)^3\right] + k\pi\ \ \ \forall x \in \mathbb{R}

[/tex]

Anyway, according to the official results, it's not correct...

Can you see the problem?

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# Homework Help: Trying to solve y'(2 - e^x) = -3e^x sin(y)cos(y)

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