Separable ODE Solution: Solving for y in dy/dt = (ab - c(y^2)) / a

[flaconlx]

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.

 

[chanvincent]

So you are trying to solve the following integration??

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

one of the very standard way to do it is to use trig substitution... i.e. y = C' sin \theta determine what C' shall you use in order to cancel out the denumerater...

 

[flaconlx]

Sorry, I must be missing something.. the integration that I was attempting to do was:

\int \frac{dy}{ab - c(y^2)}

Will trig substitution also work with this?

 

[chanvincent]

Yes... But if you use y = C' tan\theta, it might be easier to do...
just try it and see where you go...

 

[flaconlx]

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...

 

[chanvincent]

So, you are saying t = tan^{-1} (C'' y) + C''' is too complicated?
I don't think you could simplify this further...

 

[flaconlx]

how do I represent the equation in terms of y?

btw, thanks for the help so far

 

[chanvincent]

\int dt = \int \frac{1}{b}\frac{dy}{1-c/ab y^2}
t + C_0 = \frac {1}{2b} \int \frac{1}{1-\sqrt{c/ab}y} + \frac{1}{1+\sqrt{c/ab}y} dy
use z = \sqrt{c/ab}y
t + C_0 = \frac{1}{2b} \sqrt{ab/c}( ln(1+z) - ln(1-z) + C_1)
t + C_2 = 1/2\sqrt{a/bc} ln (1+z)/(1-z)
e^{2\sqrt{bc/a}t} + C_3 = (1+z)/(1-z)
z = (e^{2\sqrt{bc/a}t} + C_3 -1 ) / (e^{2\sqrt{bc/a}t} + C_3 + 1)
y = \sqrt{ab/c}(e^{2\sqrt{bc/a}t} + C_3 -1 ) / (e^{2\sqrt{bc/a}t} + C_3 + 1)

 

[flaconlx]

That actually helps me quite a bit.
Thank you!