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Thread: Homogeneous 1st order Differential Equation

  1. #1
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    Homogeneous 1st order Differential Equation

    Hi guys, I've gotten so far with this but I'm stuck on what to do next:

    Have I done the steps correctly? I followed what I was told but maybe I've made a mistake I cannot see.

    $\displaystyle (2x^2+y^2)y'=2xy$

    $\displaystyle y'= \frac{2(y/x)}{2+(y/x)^2}$

    So by letting v=y/x and differentiating and substituting my value for v and dy/dx in I get the following seperable DE

    $\displaystyle \int\frac{(-2+v^2)}{v^3} dv = \int(1/x) dx$

    Which I work out to be

    $\displaystyle \frac{1}{v^2} -\ln v = \ln x +C$

    But Im unsure how to isolate v, I tried using maple and It gave me an answer using the Lambert W function which I've never seen before, so I'm sure I've made a mistake along the lines..

    Any help is appreciated!
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  2. #2
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    Quote Originally Posted by dtwazere View Post
    Hi guys, I've gotten so far with this but I'm stuck on what to do next:

    Have I done the steps correctly? I followed what I was told but maybe I've made a mistake I cannot see.

    $\displaystyle (2x^2+y^2)y'=2xy$

    $\displaystyle y'= \frac{2(y/x)}{2+(y/x)^2}$

    So by letting v=y/x and differentiating and substituting my value for v and dy/dx in I get the following seperable DE

    $\displaystyle \int\frac{(-2+v^2)}{v^3} dv = \int(1/x) dx$

    Which I work out to be

    $\displaystyle \frac{1}{v^2} -\ln v = \ln x +C$

    But Im unsure how to isolate v, I tried using maple and It gave me an answer using the Lambert W function which I've never seen before, so I'm sure I've made a mistake along the lines..

    Any help is appreciated!
    Why not just back-substitute v = y/x and use a basic log rule to simplify ....? By the way, I think the -2 should be just 2.
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  3. #3
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    Hmm. Standard form (at least how I learned it) is

    $\displaystyle -2xy\,dx+(2x^{2}+y^{2})\,dy=0.$

    Substitute $\displaystyle x=uy,$ with $\displaystyle dx=u\,dy+y\,du,$ giving us

    $\displaystyle -2uy^{2}(u\,dy+y\,du)+(2u^{2}y^{2}+y^{2})\,dy=0,$ or

    $\displaystyle -2uy\,du+dy=0.$ Separating yields

    $\displaystyle -2u\,du+\dfrac{dy}{y}=0.$

    Can you finish? I was able to solve for $\displaystyle x,$ but not for $\displaystyle y.$
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  4. #4
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    $\displaystyle \displaystyle (2x^2 + y^2)\,\frac{dy}{dx} = 2x\,y$

    $\displaystyle \displaystyle \frac{2x^2 + y^2}{2x\,y} = \frac{dx}{dy}$

    $\displaystyle \displaystyle \frac{x}{y} + \frac{y}{2x} = \frac{dx}{dy}$

    $\displaystyle \displaystyle \frac{x}{y} + \frac{1}{2}\left(\frac{x}{y}\right)^{-1} = \frac{dx}{dy}$.

    Now make the substitution $\displaystyle \displaystyle v = \frac{x}{y} \implies x = v\,y \implies \frac{dx}{dy} = y\,\frac{dv}{dy} + v$ and the integral becomes

    $\displaystyle \displaystyle v + \frac{1}{2}v^{-1} = y\,\frac{dv}{dy} + v$

    $\displaystyle \displaystyle \frac{1}{2}v^{-1} = y\,\frac{dv}{dy}$

    $\displaystyle \displaystyle \frac{1}{2}y^{-1} = v\,\frac{dv}{dy}$

    $\displaystyle \displaystyle \int{\frac{1}{2}y^{-1}\,dy} = \int{v\,\frac{dv}{dy}\,dy}$

    $\displaystyle \displaystyle \frac{1}{2}\ln{|y|} + C_1 = \int{v\,dv}$

    $\displaystyle \displaystyle \frac{1}{2}\ln{|y|} + C_1 = \frac{1}{2}v^2 + C_2$

    $\displaystyle \displaystyle v^2 = \ln{|y|} + C$ where $\displaystyle \displaystyle C = 2(C_1 - C_2)$

    $\displaystyle \displaystyle \left(\frac{x}{y}\right)^2 = \ln{|y|} + C$

    $\displaystyle \displaystyle \frac{x^2}{y^2} = \ln{|y|} + C$

    $\displaystyle \displaystyle x^2 = y^2(\ln{|y|} + C)$

    $\displaystyle \displaystyle x = \pm y \sqrt{\ln{|y|} + C }$.


    You will not be able to isolate $\displaystyle \displaystyle y$.
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  5. #5
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    implicit

    Thanks for the help!!

    I forgot to add it says in the question to find the implicit solution,

    Would I be able to isolate it for x and then set it equal to a constant?
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  6. #6
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    Quote Originally Posted by dtwazere View Post
    Thanks for the help!!

    I forgot to add it says in the question to find the implicit solution,

    Would I be able to isolate it for x and then set it equal to a constant?
    What is "it" that you're setting equal to a constant? And why would you want to set it equal to a constant? Neither x nor y is, in this problem, in general, equal to a constant.
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