nonlinear first order DE


by utterfly
Tags: nonlinear, order
utterfly
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#1
Jul10-08, 10:01 AM
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.
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jeffreydk
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#2
Jul10-08, 04:37 PM
P: 136
Are you sure it isn't homogeneous?
utterfly
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#3
Jul10-08, 05:05 PM
P: 11
Quote Quote by jeffreydk View Post
Are you sure it isn't homogeneous?
Hi Jeffrey
The equation is not homogeneous. See if you can find a work around.
Thanks

tiny-tim
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#4
Jul10-08, 05:52 PM
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nonlinear first order DE


Quote Quote by utterfly View Post
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.
utterfly
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#5
Jul11-08, 09:04 AM
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Quote Quote by tiny-tim View Post
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
tiny-tim
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#6
Jul11-08, 09:10 AM
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Factor it out first!
utterfly
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#7
Jul11-08, 05:45 PM
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Quote Quote by tiny-tim View Post
Factor it out first!
Hi tiny-tim
I have tried factoring the equation - but no luck!
Can you help with the factoring?
Thanks
tiny-tim
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#8
Jul11-08, 06:14 PM
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Quote Quote by utterfly View Post
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)

What's difficult about that?

Now make the obvious substitution
utterfly
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#9
Jul12-08, 08:29 AM
P: 11
Quote Quote by tiny-tim View Post
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.
smallphi
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#10
Jul12-08, 10:22 AM
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.
tiny-tim
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#11
Jul12-08, 12:24 PM
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Quote Quote by smallphi View Post
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 Quote by utterfly View Post
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?
arildno
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#12
Jul12-08, 12:45 PM
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To give you a few further hints:
xy=z/x, and y/x=z/(x^3).
utterfly
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#13
Jul12-08, 05:03 PM
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!
tiny-tim
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Jul12-08, 07:03 PM
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Quote Quote by utterfly View Post
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?
utterfly
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#15
Jul13-08, 08:16 AM
P: 11
Quote Quote by tiny-tim View Post
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.
tiny-tim
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Jul13-08, 09:01 AM
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Quote Quote by utterfly View Post
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.
utterfly
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#17
Jul13-08, 10:09 AM
P: 11
Quote Quote by tiny-tim View Post
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.
tiny-tim
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#18
Jul13-08, 10:21 AM
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Quote Quote by utterfly View Post
If you have a simpler expression I would like to see it.
I won't do it for you!

But if you'd like to show your whole calculation


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