How can a first order nonlinear differential equation be solved?

In summary, the conversation discusses the process of solving a first order nonlinear differential equation that is not exact nor homogeneous. The experts suggest factoring and making a substitution in order to simplify the equation. The conversation also includes a calculation with multiple steps and a proposed solution for the equation. The solution is related to the density and radius of a gas of gravitational waves, and it may provide insight into the source of "dark energy" in the Universe.
  • #1
utterfly
11
0
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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  • #2
Are you sure it isn't homogeneous?
 
  • #3
jeffreydk said:
Are you sure it isn't homogeneous?

Hi Jeffrey
The equation is not homogeneous. See if you can find a work around.
Thanks
 
  • #4
Welcome to PF!

utterfly said:
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:
 
  • #5


tiny-tim said:
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
 
  • #6
Factor it out first! :smile:
 
  • #7
tiny-tim said:
Factor it out first! :smile:

Hi tiny-tim
I have tried factoring the equation - but no luck!
Can you help with the factoring?
Thanks
 
  • #8
factoring:

utterfly said:
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:
 
  • #9


tiny-tim said:
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.
 
  • #10
I think he meant to substitute new variable x^2 y(x) = f(x) and then separation of variables x and f works.
 
  • #11
now substitute …

smallphi said:
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:)
utterfly said:
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:
 
  • #12
To give you a few further hints:
xy=z/x, and y/x=z/(x^3).
 
  • #13
I tried the substitution; I do get a result even if looks horrible!
I will repeat the calculation, just to make sure.
Thanks guys!
 
  • #14
utterfly said:
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:
 
  • #15
tiny-tim said:
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.
 
  • #16
utterfly said:
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:
 
  • #17
tiny-tim said:
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.
 
  • #18
utterfly said:
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:
 
  • #19
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.
 
  • #20
utterfly said:
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:
 
  • #21
tiny-tim said:
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.
 
  • #22
oh I see … it was dy/dx.
utterfly said:
(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:​
 
  • #23
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
 

1. What is a nonlinear first order differential equation?

A nonlinear first order differential equation is an equation that involves an unknown function and its derivatives, where the function is not linear with respect to the independent variable. This means that the equation cannot be written in the form of y = mx + b, where m and b are constants.

2. How is a nonlinear first order differential equation different from a linear one?

A linear first order differential equation can be solved using simple algebraic methods, while a nonlinear equation requires more advanced techniques such as separation of variables or substitution to solve. Additionally, the solution to a nonlinear equation may not be a straight line, as is the case with linear equations.

3. Can a nonlinear first order differential equation have multiple solutions?

Yes, a nonlinear first order differential equation can have multiple solutions. This is because the equation is not uniquely defined by its initial conditions, unlike linear equations which have a unique solution.

4. What applications use nonlinear first order differential equations?

Nonlinear first order differential equations have numerous applications in physics, engineering, economics, and other fields. They are commonly used to model systems with complex behavior, such as population dynamics, chemical reactions, and electrical circuits.

5. How can I solve a nonlinear first order differential equation?

There is no single method for solving all nonlinear first order differential equations, as the approach will depend on the specific equation and initial conditions. However, some common techniques include separation of variables, substitution, and using integrating factors. It may also be helpful to use computer software or numerical methods to approximate a solution.

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