# Nonlinear first order DE

by utterfly
Tags: nonlinear, order
 P: 11 Hello: I discovered this forum while looking for advice on solving a first order nonlinear differential equation. The equation I am trying to solve is dy/dx=(3ay+3bx^2y^2)/(3x-bx^3y) a and b are constants. The equation is not exact, nor is it homogeneous. I have failed to separate the variables by factoring. So the usual methods don't work. Any help or advice will be appreciated.
 P: 135 Are you sure it isn't homogeneous?
P: 11
 Quote by jeffreydk Are you sure it isn't homogeneous?
Hi Jeffrey
The equation is not homogeneous. See if you can find a work around.
Thanks

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P: 26,148
Nonlinear first order DE

 Quote by utterfly dy/dx=(3ay+3bx^2y^2)/(3x-bx^3y)
Hi utterfly! Welcome to PF!

(are you sure it isn't (3ax-bx^3y) on the bottom? anyway …)

Hint: first, factor it out as much as you can, then make the obvious substitution.
P: 11
 Quote by tiny-tim Hi utterfly! Welcome to PF! (are you sure it isn't (3ax-bx^3y) on the bottom? anyway …) Hint: first, factor it out as much as you can, then make the obvious substitution.
Hello tiny-tim:
It is 3x and not 3ax.
I am going to try an iterative approach. Nothing else seems to work.
Thanks
 Sci Advisor HW Helper Thanks P: 26,148 Factor it out first!
P: 11
 Quote by tiny-tim Factor it out first!
Hi tiny-tim
I have tried factoring the equation - but no luck!
Can you help with the factoring?
Thanks
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P: 26,148
 Quote by utterfly Hi tiny-tim I have tried factoring the equation - but no luck! Can you help with the factoring? Thanks
dy/dx = (3ay + 3bx2y2)/(3x - bx3y)

= (3y/x)(a + bx2y)/(3 - bx2y)

Now make the obvious substitution …
P: 11
 Quote by tiny-tim dy/dx = (3ay + 3bx2y2)/(3x - bx3y) = (3y/x)(a + bx2y)/(3 - bx2y) What's difficult about that? Now make the obvious substitution …
Yes, you can even make it simpler by dividing through by x^2. The terms remaining have both variables. Separation of variables has not been successful.
So substitution will not help.
 P: 443 I think he meant to substitute new variable x^2 y(x) = f(x) and then separation of variables x and f works.
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P: 26,148
 Quote by smallphi I think he meant to substitute new variable x^2 y(x) = f(x) and then separation of variables x and f works.
Hi smallphi!

Exactly!

(btw, have you noticed the new x2 and x2 tags on the Reply to thread page? )
 Quote by utterfly Yes, you can even make it simpler by dividing through by x^2.
Not what I call simpler.
 The terms remaining have both variables. Separation of variables has not been successful.
Simplification is always the correct first step.

But obviously it doesn't actually solve the problem.

In hindsight, what part of "Now make the obvious substitution" did you not think worth trying?

Anyway, as smallphi suggests, put z = x2y … what is dz/dx?
 Sci Advisor HW Helper PF Gold P: 12,016 To give you a few further hints: xy=z/x, and y/x=z/(x^3).
 P: 11 I tried the substitution; I do get a result even if looks horrible! I will repeat the calculation, just to make sure. Thanks guys!
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P: 26,148
 Quote by utterfly I tried the substitution; I do get a result even if looks horrible!
Hi utterfly!

It shouldn't look horrible.

What dz/dx did you get?
P: 11
 Quote by tiny-tim Hi utterfly! It shouldn't look horrible. What dz/dx did you get?
Hi tiny-tim
This is what I get

dz/dx=(z/x)((3a+b)+z(3-2b)/(3-bz))

Solving for z gives a bunch of ln terms.
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P: 26,148
 Quote by utterfly This is what I get dz/dx=(z/x)((3a+b)+z(3-2b)/(3-bz))
hmm … that's not what I get.
 Solving for z gives a bunch of ln terms.
If you mean a sum of ln terms, then just antilog the whole thing, and you'll get a neat product of terms, without logs.
P: 11
 Quote by tiny-tim hmm … that's not what I get. If you mean a sum of ln terms, then just antilog the whole thing, and you'll get a neat product of terms, without logs.
That is true. Taking anti-logs gives the horrible expression I was referring to.
But, it is a solution!
If you have a simpler expression I would like to see it.
Much appreciate your interest and effort.