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Separable ODE Problem

  • Thread starter flaconlx
  • Start date
5
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Hi, I'm having trouble with this ODE problem:

The prompt asks to find the solution of the ODE using separation of variables.

dy/dt = (ab - c(y^2)) / a,

where a, b and c are constants.

I proceed to divide both sides by (ab - c(y^2)) and multiply both sides by dt, but I'm having trouble integrating and expressing the solution in terms of y.

Thanks in advance.
 

Answers and Replies

So you are trying to solve the following integration??

[tex] \int \frac{dy}{b - c/ay^2} [/tex]

one of the very standard way to do it is to use trig substitution.... i.e. [tex] y = C' sin \theta [/tex] determine what C' shall you use in order to cancel out the denumerater...
 
5
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Sorry, I must be missing something.. the integration that I was attempting to do was:

[tex] \int \frac{dy}{ab - c(y^2)} [/tex]

Will trig substitution also work with this?
 
Yes... But if you use [tex]y = C' tan\theta [/tex], it might be easier to do....
just try it and see where you go....
 
5
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ArcTanh[(Sqrt[c]*x)/(Sqrt[a]*Sqrt)]/ (Sqrt[a]*Sqrt*Sqrt[c])

is the solution to the integral, but it seems too complex for this problem. I guess a better question would be: is there another way to tackle the ODE so that I will arrive at an easier solution?

i.e. I need a formula in terms of y...
 
So, you are saying [tex] t = tan^{-1} (C'' y) + C'''[/tex] is too complicated???
I don't think you could simplify this further...
 
5
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how do I represent the equation in terms of y?

btw, thanks for the help so far
 
[tex] \int dt = \int \frac{1}{b}\frac{dy}{1-c/ab y^2} [/tex]
[tex] t + C_0 = \frac {1}{2b} \int \frac{1}{1-\sqrt{c/ab}y} + \frac{1}{1+\sqrt{c/ab}y} dy [/tex]
use [tex] z = \sqrt{c/ab}y [/tex]
[tex] t + C_0 = \frac{1}{2b} \sqrt{ab/c}( ln(1+z) - ln(1-z) + C_1)[/tex]
[tex] t + C_2 = 1/2\sqrt{a/bc} ln (1+z)/(1-z) [/tex]
[tex] e^{2\sqrt{bc/a}t} + C_3 = (1+z)/(1-z) [/tex]
[tex] z = (e^{2\sqrt{bc/a}t} + C_3 -1 ) / (e^{2\sqrt{bc/a}t} + C_3 + 1) [/tex]
[tex] y = \sqrt{ab/c}(e^{2\sqrt{bc/a}t} + C_3 -1 ) / (e^{2\sqrt{bc/a}t} + C_3 + 1) [/tex]
 
Last edited:
5
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That actually helps me quite a bit.
Thank you!
 

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