# Math Help - Solve the following ODE .. #7

1. ## Solve the following ODE .. #7

Hello

problem:

Solve : $y^2(1-x^2)dx+x(x^2y+2x+y)dy=0$
It must be solved using inspection
which is $ydx+xdy=d(xy)$ , $ydx-xdy=y^2d\left(\dfrac{x}{y}\right)$ ... etc

But this equation is bothering me
I tried as much I can, and it seems impossible!
any help??

2. y = 0 is a perfectly good solution.

3. I did not say that I want a particular solution.

4. You would probably need an integrating factor.

5. I can find this.
But the problem is that it should be solved using inspection.

6. Well, aside from the trivial solution that I gave you before, nothing is obvious. It's not linear in either x or y, it's not exact, homogeneous, or separable. And I don't see any obvious substitution. All of this means that, unless you can manipulate the equation to the point where you can do a direct integration, you may not be able to find a general solution at all. I certainly see no possibility for a solution by inspection, except for what I have already given.

I tried a quotient rule with a function of two variables thus: make

$\dfrac{d}{dx}\left[\dfrac{g(x,y)}{f(x)}\right]=0.$

Carrying out the differentiation and comparing to the original DE yields the two equations

$f(x)\,g_{y}(x,y)=x(x^{2}y+2x+y),$ and

$f(x)\,g_{x}(x,y)-g(x,y)\,f'(x)=y^{2}(1-x^{2}).$

Unfortunately, it's impossible to make those two equations work, as far as I can tell. You're close, actually, very close. I get

$f(x)=x,$

$g(x,y)=\dfrac{x^{2}y^{2}}{2}+2xy+\dfrac{y^{2}}{2}.$

But plugging this into the second equation above yields

$\dfrac{x^{2}y^{2}}{2}-\dfrac{y^{2}}{2}=y^{2}(1-x^{2}),$

which is off by a minus sign and a fraction of 1/2.

This approach might be made to work, but I'm not seeing how.

7. $
y^2(1-x^2)dx+x(x^2y+2x+y)dy=0
$

$
y^2dx-x^2y^2dx+x^3ydy+2x^2dy+xydy=0
$

Re-arrange:

$
y^2dx+xydy+2x^2dy+x^3ydy-x^2y^2dx=0
$

Taking common factors:

$
y(ydx+xdy)+2x^2dy+x^2y(xdy-ydx)=0
$

Or:

$
yd(xy)+2x^2dy+x^4y \, d\left(\dfrac{y}{x}\right)=0
$

Devide by $x^2y^3$ :

$
\dfrac{d(xy)}{(xy)^2}+2 \, \dfrac{dy}{y^3} + \dfrac{x^2}{y^2} \, d\left(\dfrac{y}{x}\right) = 0
$

Or:

$
\dfrac{d(xy)}{(xy)^2}+2 \, \dfrac{dy}{y^3} + \dfrac{1}{\left(\dfrac{y}{x}\right)^2} \, d\left(\dfrac{y}{x}\right) = 0
$

Integrate.

8. Originally Posted by General
$
y^2(1-x^2)dx+x(x^2y+2x+y)dy=0
$

$
y^2dx-x^2y^2dx+x^3ydy+2x^2dy+xydy=0
$

Re-arrange:

$
y^2dx+xydy+2x^2dy+x^3ydy-x^2y^2dx=0
$

Taking common factors:

$
y(ydx+xdy)+2x^2dy+x^2y(xdy-ydx)=0
$

Or:

$
yd(xy)+2x^2dy+x^4y \, d\left(\dfrac{y}{x}\right)=0
$

Devide by $x^2y^3$ :

$
\dfrac{d(xy)}{(xy)^2}+2 \, \dfrac{dy}{y^3} + \dfrac{x^2}{y^2} \, d\left(\dfrac{y}{x}\right) = 0
$

Or:

$
\dfrac{d(xy)}{(xy)^2}+2 \, \dfrac{dy}{y^3} + \dfrac{1}{\left(\dfrac{y}{x}\right)^2} \, d\left(\dfrac{y}{x}\right) = 0
$

Integrate.
How in tarnation did you ever see to do that?

9. Practice makes perfect =)

10. I guess so! Incidentally, integration of your brilliant equation there yields a 1-parameter family of solutions, for which the solution y = 0 is a singular solution.

Cheers.

11. Thanks !!

12. Well, you're certainly welcome for my contribution, but you can obviously see that General did the heavy lifting.

13. Actually, the "thanks" for General not for you xD
but thank you anyway

14. Originally Posted by Liverpool
Actually, the "thanks" for General not for you xD
but thank you anyway
Perfectly fine. Like I said, General did the heavy lifting.

15. I See.
Thanks all.