- #1
twoflower
- 368
- 0
Hi,
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?
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?