# Solve the following ODE .. #7

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• Feb 9th 2011, 03:25 AM
Liverpool
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??
• Feb 9th 2011, 04:48 AM
Ackbeet
y = 0 is a perfectly good solution.
• Feb 9th 2011, 05:16 AM
Liverpool
I did not say that I want a particular solution.
• Feb 9th 2011, 05:37 AM
Prove It
You would probably need an integrating factor.
• Feb 9th 2011, 05:42 AM
Liverpool
I can find this.
But the problem is that it should be solved using inspection.
• Feb 9th 2011, 07:06 AM
Ackbeet
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.
• Feb 9th 2011, 08:41 AM
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.
• Feb 9th 2011, 08:44 AM
Ackbeet
Quote:

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?
• Feb 9th 2011, 08:49 AM
General
Practice makes perfect =)
• Feb 9th 2011, 08:51 AM
Ackbeet
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.
• Feb 11th 2011, 11:06 AM
Liverpool
Thanks !!
• Feb 11th 2011, 11:14 AM
Ackbeet
Well, you're certainly welcome for my contribution, but you can obviously see that General did the heavy lifting.
• Feb 11th 2011, 11:54 AM
Liverpool
Actually, the "thanks" for General not for you xD
but thank you anyway :p
• Feb 11th 2011, 11:57 AM
Ackbeet
Quote:

Originally Posted by Liverpool
Actually, the "thanks" for General not for you xD
but thank you anyway :p

Perfectly fine. Like I said, General did the heavy lifting.
• Feb 11th 2011, 11:58 AM
Liverpool
I See.
Thanks all.