nonlinear first order DE

[utterfly]

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.

[jeffreydk]

Are you sure it isn't homogeneous?

[utterfly]

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]

dy/dx=(3ay+3bx^2y^2)/(3x-bx^3y)

Hi utterfly! Welcome to PF! :smile:

(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. :smile:

[utterfly]

Hi utterfly! Welcome to PF! :smile:

(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. :smile:
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]

Factor it out first! :smile:

[utterfly]

Factor it out first! :smile:

Hi tiny-tim
I have tried factoring the equation - but no luck!
Can you help with the factoring?
Thanks

[tiny-tim]

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? :confused:

Now make the obvious substitution … :smile:

[utterfly]

dy/dx = (3ay + 3bx2y2)/(3x - bx3y)

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

What's difficult about that? :confused:

Now make the obvious substitution … :smile:

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]

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]

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! :smile:

Exactly! :biggrin:

(btw, have you noticed the new x2 and x2 tags on the Reply to thread page? :smile:)
Yes, you can even make it simpler by dividing through by x^2.

Not what I call simpler. :confused:
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? :smile:

[arildno]

To give you a few further hints:
xy=z/x, and y/x=z/(x^3).

[utterfly]

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]

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? :smile:

[utterfly]

Hi utterfly!

It shouldn't look horrible.

What dz/dx did you get? :smile:
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]

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. :smile:

[utterfly]

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. :smile:

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]

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 … :smile:

[utterfly]

The calculation is lengthy. Here goes: f=x^2y
(1/x^2)df/dx-2f/x^3=3f/x^3((a+f)/(3-bf))
df/dx=(f/x)((3a+b)+f(3-2b))/(3-bf))
Int((3-bf)/((3a+b)+f(3-2b))df/f=Int(dx/x)
(1/(a-2))lnf-(3a/2(3a-6))ln((3a-b)+f(3-2b))=lnx+K
The ln terms with f combine into two terms. lnf=2lnx+lny is substituted.
When the ln terms are eliminated by raising to a power, the final expression for y(x) is transcendental.

[tiny-tim]

The calculation is lengthy. Here goes: f=x^2y
(1/x^2)df/dx-2f/x^3=3f/x^3((a+f)/(3-bf)) …

oooh … on the LHS, you've differentiated as if f = x/y2. :cry:

(btw, do try using the X2 and X2 tags on the Reply page)

Try again! :smile:

[utterfly]

oooh … on the LHS, you've differentiated as if f = x/y2. :cry:

(btw, do try using the X2 and X2 tags on the Reply page)

Try again! :smile:
f=x^2 y
y=f/x^2
dy/dx= -2f/x^3 + (1/x^2)df/dx
This seems OK

Sorry can't get the sub and sup to work.

[tiny-tim]

oh I see … it was dy/dx.
(1/x^2)df/dx-2f/x^3=3f/x^3((a+f)/(3-bf))
df/dx=(f/x)((3a+b)+f(3-2b))/(3-bf))

At the end of the first line, shouldn't it be (a+bf)/(3-bf)?

And I don't see how you got the next line. :confused:

Going to bed now … :zzz:

[utterfly]

I simplified the solution to a quadratic equation. It looks OK, except that the constant of integration is attached to 'x' inside the radical.
I will make use of it.
If you are curious: 'y' is the density, 'x' is the radius of a gas of gravitational waves. The solution relates the density with radius - its equation of state.
The solution will go into the source term of the Einstein equation. The goal is to see if the evolution of the Universe is driven by gravitational waves emitted at the Big Bang. The solution will be fitted to supernova data from which the constant will be computed. The constant is the total energy of the Universe.
Maybe gravitational waves of cosmological origin is the source of "dark energy".
Many thanks