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Non-linear ODE.

  1. Apr 13, 2008 #1
    I'm curious about the differential equation which takes the general form of,
    Where C and D are constants. According to mathematica, the answer is:
    [tex] \sqrt{\frac{D}{C}} \tanh{(x \sqrt{CD})}[/tex].
    But I'd like to know how this is done by hand, I was able to do this with separation of variables and finding x, then solving for y, but it was a large waste of time and I'm curious if there is a general way to solve this just as linear ones can be solved.
  2. jcsd
  3. Apr 13, 2008 #2


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    no that is why linear equations are nice they are much easier to solve and can always (at least in principle) be solved. It is a simple integration.
  4. Apr 22, 2008 #3
  5. Apr 24, 2008 #4


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    I don't know why you would use the Riccatti equation. This is a simple, separable, first order differential equation. It can be integrated directly. I don't know exactly what gamesguru did or why he considers it a "large waste of time" but it is not difficult. It separates into
    [tex]\frac{dy}{Cy^2+ D}= dx[/itex]
    multiplying on both sides by D, we get
    [tex]\frac{dy}{\frac{C}{D}y^2+ 1}= Ddx[/tex]
    Now, what we do depends upon the signs of C and D. If C and D have opposite signs so that C/D is negative, write the equation as
    [tex]\frac{dy}{1- \left|\frac{C}{D}}\right|y^2}= Ddx[/tex]
    and let [itex]u= \sqrt{|C/D|}[/itex]. Then the equation becomes
    [tex]\sqrt{|C/D|}\frac{du}{u^2+ 1}= Ddt[/tex]
    and the solution is exactly what gamesguru said.

    If C and D have the same sign, let u= [itex]\sqrt{C/D}y[/itex] so that the equation becomes
    [tex]\sqrt{\frac{D}{C}}\frac{du}{u^2+ 1}= Ddt[/tex]
    and integrating gives the same thing with "tan" rather than "tanh".
  6. Apr 24, 2008 #5


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    Shouldn't it be:
    \frac{dy}{D-Cy^2}= dx
    The answer changes, but the method is the same.
  7. Apr 24, 2008 #6
    Yes, but he wanted to know if there was a way to do it without separation of variables.
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