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Math Help - Solve the following ODE .. #7

  1. #1
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    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??
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  2. #2
    A Plied Mathematician
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    y = 0 is a perfectly good solution.
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  3. #3
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    I did not say that I want a particular solution.
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  4. #4
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    You would probably need an integrating factor.
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  5. #5
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    I can find this.
    But the problem is that it should be solved using inspection.
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  6. #6
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    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.
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  7. #7
    Super Member General's Avatar
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    <br />
y^2(1-x^2)dx+x(x^2y+2x+y)dy=0<br />

    <br />
y^2dx-x^2y^2dx+x^3ydy+2x^2dy+xydy=0<br />

    Re-arrange:

    <br />
y^2dx+xydy+2x^2dy+x^3ydy-x^2y^2dx=0<br />

    Taking common factors:

    <br />
y(ydx+xdy)+2x^2dy+x^2y(xdy-ydx)=0<br />

    Or:

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

    Devide by x^2y^3 :

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

    Or:

    <br />
\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<br />

    Integrate.
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  8. #8
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    Quote Originally Posted by General View Post
    <br />
y^2(1-x^2)dx+x(x^2y+2x+y)dy=0<br />

    <br />
y^2dx-x^2y^2dx+x^3ydy+2x^2dy+xydy=0<br />

    Re-arrange:

    <br />
y^2dx+xydy+2x^2dy+x^3ydy-x^2y^2dx=0<br />

    Taking common factors:

    <br />
y(ydx+xdy)+2x^2dy+x^2y(xdy-ydx)=0<br />

    Or:

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

    Devide by x^2y^3 :

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

    Or:

    <br />
\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<br />

    Integrate.
    How in tarnation did you ever see to do that?
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  9. #9
    Super Member General's Avatar
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    Practice makes perfect =)
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  10. #10
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    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.
    Last edited by Ackbeet; February 9th 2011 at 09:01 AM.
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  11. #11
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    Thanks !!
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  12. #12
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    Well, you're certainly welcome for my contribution, but you can obviously see that General did the heavy lifting.
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  13. #13
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    Actually, the "thanks" for General not for you xD
    but thank you anyway
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  14. #14
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    Quote Originally Posted by Liverpool View Post
    Actually, the "thanks" for General not for you xD
    but thank you anyway
    Perfectly fine. Like I said, General did the heavy lifting.
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  15. #15
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    I See.
    Thanks all.
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