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Thread: Solve first order ODE

  1. #1
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    Solve first order ODE

    The ODE is
    $xdx+ydy=\frac {a^2 (xdy-ydx)}{x^2+y^2} $

    We can take $x^2+y^2$ on the left hand side to get
    $(x^2+y^2)d (x^2+y^2)=2a^2 (xdy-ydx) $

    I am not able to move ahead from here.
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  2. #2
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    Re: Solve first order ODE

    The first thing to do would be to gather all the similar differentials together.

    \big((x^2+y^2)x+2a^2y)\mathrm dx + \big((x^2+y^2)y-2a^2x\big)\mathrm dy)=0

    I would then test to see if it is an exact equation. If it isn't, seek an integrating factor.
    Last edited by Archie; Sep 27th 2016 at 05:33 AM.
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  3. #3
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    Re: Solve first order ODE

    I have tried doing this. The problem here is that the DE is not exact and you can only use the integration factor 1/bx+ay where b is the eqn before dx and a is eqn before dy.The integration factor results in an equation with more than 1 term which makes figuring out the differential very unobvious and time consuming.
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  4. #4
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    Re: Solve first order ODE

    On the contrary, the equation you get by multiplying both sides by x^2+ y^2 is exact.

    (x^2+ y^2)(xdx+ ydy)= a^2(xdy+ ydx)
    (x^3+ xy^2)dx+ (x^2y+ y^3)dy= a^2xdy+ a^2ydx
    (x^3+ xy^2- a^2y)dx+ (x^2y+ y^3- a^2x)dy= 0

    The derivative of x^3+ xy^2- a^2y with respect to y is 2xy- a^2 and the derivative of x^2y+ y^3- a^2x with respect to x is 2xy- a^2. Those are the same so the equation is exact.
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  5. #5
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    Re: Solve first order ODE

    You have made an error in the DE
    To the right of the equality we have $a^2 (xdy-ydx)$
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  6. #6
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    Re: Solve first order ODE

    .
    Last edited by Archie; Sep 27th 2016 at 10:43 AM.
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  7. #7
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    Re: Solve first order ODE

    Quote Originally Posted by sahigg View Post
    You have made an error in the DE
    To the right of the equality we have $a^2 (xdy-ydx)$
    Oh, darn!
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  8. #8
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    Re: Solve first order ODE

    Quote Originally Posted by sahigg View Post
    The ODE is
    $xdx+ydy=\frac {a^2 (xdy-ydx)}{x^2+y^2} $

    We can take $x^2+y^2$ on the left hand side to get
    $(x^2+y^2)d (x^2+y^2)=2a^2 (xdy-ydx) $

    I am not able to move ahead from here.
    $u=x^2 + y^2$

    $v = \dfrac x y$

    $\dfrac{du}{dx}=2x + 2y \dfrac{dy}{dx}$

    $du = 2x dx + 2y dy$

    $x dx + y dy = \dfrac 1 2 du$

    $\dfrac{dv}{dx} = \dfrac{y - x \frac{dy}{dx}}{y^2}$

    $y^2 dv = y dx - x dy$

    $\dfrac{u}{1+v^2}dv = ydx - x dy$

    $a^2 \dfrac{x dy - y dx}{x^2+y^2} = -a^2 \dfrac{dv}{1+v^2}$

    Now we can rewrite the original diff eq as

    $\dfrac{du}{2} = -a^2 \dfrac{dv}{1+v^2}$

    integrating we get

    $\dfrac{u}{2} = -a^2 \arctan(v) + C$

    using the original variables and rearranging we get

    $\dfrac{x^2 + y^2}{2} +a^2 \arctan\left(\dfrac x y\right) = C$

    and this is the implicit solution.
    Thanks from sahigg
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