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Solve dy/dx=e^(3x+2y) by separation of variables

  1. May 30, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    solve dy/dx=e^(3x+2y) by separation of variables

    3. The attempt at a solution

    [itex]\frac{dy}{dx}=e^{3x+2y}[/itex]

    [itex]\frac{dy}{dx}=e^{3x}e^{2y}[/itex]

    [itex]e^{-2y}dy=e^{3x}dx[/itex]

    [itex]\int e^{-2y}dy=\int e^{3x}dx[/itex]

    [itex]e^{-2y}=-\frac{2}{3}e^3x + C[/itex]

    [itex]-2y = ln(-\frac{2}{3}e^{3x}+C)[/itex]

    [itex]y=1\frac{1}{2}ln(-\frac{2}{3}e^{3x}+C)[/itex]

    Just a little wondering mostly about the natural log thing. Can I take the natural log of the -ve function wrt x because of the constant?
     
  2. jcsd
  3. May 30, 2012 #2

    SammyS

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    (Fixed a couple of typos above.)

    Of course this gives a "family" of solutions, depending upon the value of C.

    For each value of C, the solution is real, only if [itex]\displaystyle C-\frac{2}{3}e^{3x}>0\ \text{ i.e. } x<\frac{1}{3}\ln\left(\frac{3C}{2}\right) \ .[/itex]
     
    Last edited: May 30, 2012
  4. May 31, 2012 #3

    ElijahRockers

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    so I don't need any absolute value brackets or anything? I am a little shaky when it comes to logs/abs value
     
  5. May 31, 2012 #4

    HallsofIvy

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    No, you don't need absolute value here. If [itex]e^x= y[/itex], then ln(y)= x.

    You are thinking of the integral, [itex]\int (1/x)dx= ln|x|+ C[/itex], but you are not doing that here. [itex]\int e^{ax}dx= (1/a)e^{ax}+ C without any absolute value.
     
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