# Thread: Question to solve Differential Equation ?

1. ## Question to solve Differential Equation ?

Hello everybody

I have tried to solve this DE :

(3y^3 - xy)dx - (x^2 + 6xy^2)dy = 0

by finding the integrating factor , all manners which i did have failed to solve it , could any one help me to find suitable manner to find a suitable integrating factor for it ? i will thankful for him .

Best regards
Razi

2. ## Re: Question to solve Differential Equation ?

What have you tried? Can you show some of your work so we can see if we can find something that would help?

3. ## Re: Question to solve Differential Equation ?

If you just want the answer, here it is:

https://www.wolframalpha.com/input/?...y%5E2)dy+%3D+0

4. ## Re: Question to solve Differential Equation ?

Dear SlipEternal

Thanks for your replay , please see the attached photo to see what i did trying to solve this ODE .
About the wolframalpha.com , it is give you just the final answer without the steps , i need the steps to know which method is suitable to chose the integrating factor.

Thanks in advance and best regards
Razi

5. ## Re: Question to solve Differential Equation ?

You have

$(3y^2-x)ydx-(x^2+6xy^2)dy=0$
$(3y^2-x)y-(x^2+6xy^2)y' = 0$
$\dfrac{3y^2-x}{x^2+6xy^2} = \dfrac{y'}{y}$

The goal is to integrate both sides with respect to $x$. The RHS becomes $\ln y$. We just need to worry about the LHS.

$\dfrac{3y^2-x}{x^2+6xy^2} = \dfrac{1}{2x}-\dfrac{3}{2x+12y^2}$

Let's integrate:

$\displaystyle \ln \sqrt{x} -\dfrac{3}{2}\int \dfrac{dx}{x+6y^2} = \ln y$

I believe you are going to wind up with an integration by parts type of scenario. This will yield an infinite series, but I do not have time to work it out. Perhaps another guru can help?

6. ## Re: Question to solve Differential Equation ?

I thought about this a bit more, and I may be on the wrong track. Let's work backwards, starting with the solution and seeing if we can get back to the original problem.

$y = \sqrt{\dfrac{x}{6}W\left(\dfrac{6c_1}{x^3}\right)}$

$\dfrac{6y^2}{x} = W\left(\dfrac{6c_1}{x^3}\right)$

$\dfrac{6y^2}{x}e^{(6y^2)/x} = \dfrac{6c_1}{x^3}$

$x^2y^2e^{6y^2/x} = c_1$

Differentiating both sides, we get:

$x^2y^2e^{6y^2/x}\dfrac{12xydy-6y^2dx}{x^2}+(2xy^2dx+2x^2ydy)e^{6y^2/x} = 0$

Collecting like terms, we have: $(2xy^2-6y^4)e^{6y^2/x}dx + (12xy^3+2x^2y)e^{6y^2/x}dy = 0$

Factoring gives:

$-2ye^{6y^2/x}\left[(3y^3-xy)dx-(x^2+6xy^2)dy\right] = 0$

So, the integrating factor is $-2ye^{6y^2/x}$.

I am not sure how you would get that, though. It has been far too long since I have done differential equations.

7. ## Re: Question to solve Differential Equation ?

I managed to get $\mu =\frac{1}{x^2 y}$ as an integrating factor. So $\mu M = \frac{3y^2}{x^2}-\frac{1}{x}$ and $\mu N= \frac{-1}{y}-\frac{6y}{x}$.

So that $\left( \mu M \right)_y = \frac{6y}{x^2}$ and $\left(\mu N\right)_x =-\left(\frac{-6y}{x^2}\right)=\frac{6y}{x^2}$
Not sure if this is correct though (a little rusty).

8. ## Re: Question to solve Differential Equation ?

What about the substitution $u=\dfrac{y}{x}$?

9. ## Re: Question to solve Differential Equation ?

I didn't continue with the question sorry (getting late), but my thought was to seek an integrating factor of the form $\mu (x,y) = x^m y^n$ with $m \neq 0$ and $n \neq 0$ that makes the equation exact since it cannot be expressed as $\mu (x)$ or $\mu (y)$ as OP as shown. I'm not 100% sure if that's the way to go about it.

10. ## Re: Question to solve Differential Equation ?

Now the OP has two valid answers

11. ## Re: Question to solve Differential Equation ?

Gentlemen , really I appreciate your help , finally i have found the solution of this DE by multiply it with x^a * y^b as an integrating factor and then did some processes , however , the integrating factor is : 1/(x^2*y) .

Thanks again dear friends for your help
Best regards
Razi