# Nonlinear first order DE

1. Jul 10, 2008

### 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.

2. Jul 10, 2008

### jeffreydk

Are you sure it isn't homogeneous?

3. Jul 10, 2008

### utterfly

Hi Jeffrey
The equation is not homogeneous. See if you can find a work around.
Thanks

4. Jul 10, 2008

### tiny-tim

Welcome to PF!

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.

5. Jul 11, 2008

### utterfly

Re: Welcome to PF!

Hello tiny-tim:
It is 3x and not 3ax.
I am going to try an iterative approach. Nothing else seems to work.
Thanks

6. Jul 11, 2008

### tiny-tim

Factor it out first!

7. Jul 11, 2008

### utterfly

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

8. Jul 11, 2008

### tiny-tim

factoring:

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

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

Now make the obvious substitution …

9. Jul 12, 2008

### utterfly

Re: factoring:

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. Jul 12, 2008

### smallphi

I think he meant to substitute new variable x^2 y(x) = f(x) and then separation of variables x and f works.

11. Jul 12, 2008

### tiny-tim

now substitute …

Hi smallphi!

Exactly!

(btw, have you noticed the new x2 and x2 tags on the Reply to thread page? )
Not what I call simpler.
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?

12. Jul 12, 2008

### arildno

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

13. Jul 12, 2008

### 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!

14. Jul 12, 2008

### tiny-tim

Hi utterfly!

It shouldn't look horrible.

What dz/dx did you get?

15. Jul 13, 2008

### utterfly

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. Jul 13, 2008

### 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.

17. Jul 13, 2008

### utterfly

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. Jul 13, 2008

### tiny-tim

I won't do it for you!

But if you'd like to show your whole calculation …

19. Jul 13, 2008

### 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.

20. Jul 13, 2008

### tiny-tim

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

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

Try again!