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Math Help - ODE Proof: Solution to this will always exist?

  1. #1
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    ODE Proof: Solution to this will always exist?

    Problem:
    Prove that with the aid of the substitution y=vx, you can solve any
    equation of the form:

    y^n f(x) dx + H(x,y) \, (ydx-xdy) = 0

    where H is homogeneous function in x and y.

    Solution:
    I rearranged the equation to get:

    \left(y^n f(x) + y H(x,y) \right) dx - x H(x,y) dy = 0

    Since y=vx \implies dy=vdx+xdv

    Using this in the equation:

    \left[ v^nx^n f(x) + vx H(x,vx) \right] dx - x H(x,vx) (vdx+xdv) = 0

    Rearrange :

    v^nx^n f(x)dx - x^2 H(x,vx) dv = 0

    Now, I stopped!

    The last equation is almost separable, but that H is bothering me.

    Any help?
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  2. #2
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    But your told that H(x,y) is homogeneous which means it's of the form H = H\left(\dfrac{y}{x} \right).
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  3. #3
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    so it will be H\left(\frac{vx}{x}\right)=H(v)
    and hence it is separable.
    Right?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Liverpool View Post
    so it will be H\left(\frac{vx}{x}\right)=H(v)
    and hence it is separable.
    Right?
    yup
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  5. #5
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    Thanks.
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