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Math Help - First Order Homogeneous Diff EQ

  1. #1
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    First Order Homogeneous Diff EQ

    For some reason, this is giving me fits. Im supposed to put it in the form: M(x,y)dx+N(x,y)dy=0 then use the substitution v=\frac{y}{x}, so y=vx and dy=vdx+xdv

    Here is the D.E.

    x\frac{dy}{dx}=y+\sqrt{x^2+y^2}

    Is it safe to make it like this: (x-y)dy=\sqrt{x^2+y^2}dx then \sqrt{x^2+y^2}dx-(x-y)dy=0 then use the substitution?

    Thanks
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  2. #2
    MHF Contributor Calculus26's Avatar
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    if you multiply by dx you get:

    xdy = ydx + sqrt(x^2 +y^2)dx

    (y + sqrt(x^2+y^2))dx -xdy = 0

    y =vx

    using dy = vdx + xdv => (vx + sqrt(x^2+v^2x^2)dx - x(vdx+xdv) = 0

    (vx + x sqrt(1+v^2) -xv)dx - x^2dv = 0

    (xsqrt(1+v^2))dx -x^2 dv = 0

    which now separates 1/x dx = dv/sqrt(1+v^2)


    Typically this is not how you would approach this problem--no need to put

    in the form Mdx + Ndy = 0.



    dy/dx = y/x + sqrt[1+(y/x)^2]


    simply y = vx dy/dx = v + xdv/dx

    v+ xdv/dx = v +sqrt(1+v^2)

    dv/sqrt(1+v^2) = dx/x

    By the way if dy/dx = f(y/x) we call it a differnetial equation with homogeneous coefficients which can be solved by letting y = vx
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  3. #3
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    Ahh, I see..

    I got (x\sqrt{1+v^2})dx=x^2dv then to \frac{(\sqrt{1+v^2})dx}{x}=dv then to bring the \sqrt{1+v^2} over under the dv by diving, so then you get \frac{dx}{x}=\frac{dv}{\sqrt{1+v^2}} then integrate, and solve...sweet!

    Thanks for the help.
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