# Math Help - Nonlinear nonexact first order ODE

1. ## Nonlinear nonexact first order ODE

Hey
How can I solve
dy/dx = (3x^2 * y + y^2) / (2x^3 + 3xy)
Hopefully you can understand that. It is clearly nonlinear. It is also nonexact and I cannot find an integrating factor that works. I also tried y=vx to no avail. Can someone help me out with this and point me in the right direction?
Thanks

2. ## Re: Nonlinear nonexact first order ODE

Find g(x,y) to make diff eq exact:

$x\left(2x^2+3 y\right)dx-y\left(3x^2+y\right)dy=0$

$x\left(2x^2+3 y\right)\frac{1}{x y\left(3x^2+y\right)+2y x\left(2x^2+3 y\right)}dx-y\left(3x^2+y\right)\frac{1}{x y\left(3x^2+y\right)+2y x\left(2x^2+3 y\right)}dy=0$

Then solution is:

$x^3 y+ x y^2=c$

$y\text{=}\frac{-x^3\pm \sqrt{4 c x+x^6}}{2 x}$

for c=1,2,3,....16

also check out:

$f(x,y)=C+\frac{1}{7} \left(\log \left(x^2+y\right)+\log (x)\right)+\frac{2 \log (y)}{7}$

3. ## Re: Nonlinear nonexact first order ODE

How is that supposed to help? I am not familiar with diagrams like these for differential equations at all

4. ## Re: Nonlinear nonexact first order ODE

Hi MaxJasper !
There is a mistake in your first line (permutation of dx and dy).

Hi Shanter !
Your differential equation is solvable, but very arduously (too difficult to an home work).
I suspect that there is a "-" missing in the wording :
If the equation was : dy/dx = - (3x^2 * y + y^2) / (2x^3 + 3xy) then the integrating factor would be easy to find and the result : (x^3)(y^2)+(y^3)x = C

5. ## Re: Nonlinear nonexact first order ODE

For ODE:
dy/dx = (3x^2 * y + y^2) / (2x^3 + 3x y^2)

6. ## Re: Nonlinear nonexact first order ODE

and I missed the negative, which really simplifies everything now that I can see that

7. ## Re: Nonlinear nonexact first order ODE

Originally Posted by JJacquelin
Hi Shanter !
Your differential equation is solvable, but very arduously (too difficult to an home work).
I suspect that there is a "-" missing in the wording :
If the equation was : dy/dx = - (3x^2 * y + y^2) / (2x^3 + 3xy) then the integrating factor would be easy to find and the result : (x^3)(y^2)+(y^3)x = C
Hi folks,

How did you solve the modified version with (-) sign included?

Your solution: (x^3)(y^2)+(y^3)x = C
results in :

y'=-(3 x^2 y^2 + y^3)/(2 x^3 y + 3 x y^2)

8. ## Re: Nonlinear nonexact first order ODE

Originally Posted by MaxJasper
How did you solve the modified version with (-) sign included?
Your solution: (x^3)(y^2)+(y^3)x = C
results in :
y'=-(3 x^2 y^2 + y^3)/(2 x^3 y + 3 x y^2)
Hi MaxJasper !
Yes, the solution: (x^3)(y^2)+(y^3)x = C results in :
y' = -(3 x^2 y^2 + y^3)/(2 x^3 y + 3 x y^2) = -(3x^2 * y + y^2) / (2x^3 + 3xy)
Solving the modified version with (-) is rather easy. I let Shanter answer to your question since he found how to do it.

9. ## Re: Nonlinear nonexact first order ODE

Differentiate your solution and see that it is different from original diff eq.

10. ## Re: Nonlinear nonexact first order ODE

Originally Posted by MaxJasper
Differentiate your solution and see that it is different from original diff eq.
I don't agree, the solution is allright.
Publish on tbe forum what you did and we will locate your mistake.

11. ## Re: Nonlinear nonexact first order ODE

Your solution: (x^3)(y^2)+(y^3)x = C
results in :

y'=-(3 x^2 y^2 + y^3)/(2 x^3 y + 3 x y^2)

that is different from original diff eq. Take a careful look.

12. ## Re: Nonlinear nonexact first order ODE

Why don't symplify ?
(y) is a factor of the numerator and of the denominator.
After simplification, compare to the diff. equation (the modified version with - , of course).

13. ## Re: Nonlinear nonexact first order ODE

Hi MaxJasper !

do yo agree now ?

14. ## Re: Nonlinear nonexact first order ODE

Hey this might be a little late but if you write it out as
Mdx + Ndy = 0
To find the integrating factor I think I used
(Nx - My)/M = A, int factor = e^integral A = y
Note, A must be a function of y only.
I don't have my work with me and I am typing this on my phone but I think this is all right. The rest is straight forward using everything you have.