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Math Help - Homogeneous equation

  1. #1
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    Homogeneous equation

    Show that the given equation is homogeneous and solve the differential equation.

     dy/dx=x^2+3y^2/2xy<br />

    To show that it is homogeneous, I have dy/dx=1+3(y/x)^2/2(y/x)

    I am pretty sure that this is correct so far. Then,
    x(dv/dx)+v=1+3(y/x)^2/2(y/x)
    x(dv/dx)=1+3(y/x)^2/2(y/x)-v
    x(dv/dx)=(1+3v^2/2v)-v


    It is after this where I'm not quite sure what to do. According to my book the answer is x^2+y^2-cx^3=0. Thanks for any help
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  2. #2
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    Quote Originally Posted by steph3824 View Post
    Show that the given equation is homogeneous and solve the differential equation.

     dy/dx=x^2+3y^2/2xy<br />

    To show that it is homogeneous, I have dy/dx=1+3(y/x)^2/2(y/x)

    I am pretty sure that this is correct so far. Then,
    x(dv/dx)+v=1+3(y/x)^2/2(y/x)
    x(dv/dx)=1+3(y/x)^2/2(y/x)-v
    x(dv/dx)=(1+3v^2/2v)-v


    It is after this where I'm not quite sure what to do. According to my book the answer is x^2+y^2-cx^3=0. Thanks for any help
    It helps to use brackets. So

     <br />
\frac{dy}{dx} = \frac{x^2+3y^2}{2xy}<br />
    so if y = x v then \frac{dy}{dx} = x \frac{dv}{dx} + v

    so x \frac{dv}{dx} = \frac{1 + 3v^2}{2v} - v = \frac{1+v^2}{2v} as you've said. Now separate

     <br />
\frac{2v\,dv}{1+v^2} = \frac{dx}{x}<br />
and integrate.
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  3. #3
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    Ok I integrated that and ended up with ln|1+v^2|=ln|x|+C

    Is that correct? Where do I go from here?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    exponentiate

    1+ v^2 = C x

    replace v = y/x

    simplify
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  5. #5
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    Quote Originally Posted by steph3824 View Post
    Ok I integrated that and ended up with ln|1+v^2|=ln|x|+C

    Is that correct? Where do I go from here?
    Quote Originally Posted by Calculus26 View Post
    exponentiate

    1+ v^2 = C x

    replace v = y/x

    simplify
    Just an added note. As C is constant in ln|1+v^2|=ln|x|+C, replace it with \ln C and then follow Calc26.
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