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

  1. Dec 17, 2005 #1
    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?
     
  2. jcsd
  3. Dec 17, 2005 #2
    I got it finally.
    You know, in official results there is:

    [tex]
    y_{a,b,k} = \left\{ \begin{array}{rcl}
    \arctan(a(e^{x}-2)^3) + k\pi, & x \in (-\infty, \log 2] \\
    \arctan(b(e^{x}-2)^3) + k\pi, & x \in [\log 2, \infty)\end{array}\right.
    [/tex]

    [tex]
    \mbox{for }a,\ b \in \mathbb{R} \mbox{ and } k \in \mathbb{Z}\mbox{, and }
    [/tex]

    [tex]
    y \equiv k\frac{\pi}{2},\ \ \ x \in \mathbb{R}
    [/tex]

    and I was confused why the interval is split around the [itex]\log 2[/itex] point. Now I see it's just expressing all maximal solutions as in this point the function may get different and still satisfy the equation.

    One more technical thing which seems quite tricky to me:

    In these official results, there are [itex]a, b \in \mathbb{R}[/itex] although the critical place where they come from looks like this:

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

    [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]

    Well, strictly adjusting the equation with C involved from integrating, I got [itex]e^{C}[/itex]. Anyway, putting any real constant there (and not only positive) is also ok. One must be very careful when expressing all maximum solutions :frown:
     
    Last edited: Dec 17, 2005
  4. Dec 19, 2005 #3

    CarlB

    User Avatar
    Science Advisor
    Homework Helper

    Good job. And beautiful formatting. You should be answering problems here, or are you already?

    Carl
     
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