Non-linear ODE.

  • Thread starter gamesguru
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  • #1
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Main Question or Discussion Point

I'm curious about the differential equation which takes the general form of,
[tex]y'+Cy^2=D[/tex].
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.
 

Answers and Replies

  • #2
lurflurf
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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
HallsofIvy
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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".
 
  • #5
exk
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Shouldn't it be:
[tex]
\frac{dy}{D-Cy^2}= dx
[/tex]
?
The answer changes, but the method is the same.
 
  • #6
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I don't know why you would use the Riccatti equation.
Yes, but he wanted to know if there was a way to do it without separation of variables.
 

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