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Thread: Differential Equations

  1. #1
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    Differential Equations

    Use the substiution y=xu to find the general solution of the differential equation:

    (x+y)dy/dx=y

    dy/dx = u+x(du/dx)

    (x+xu)(u+x(du/dx)=xu

    where do i go from here?

    thanks
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  2. #2
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    To guide you, why are you making that substitution? (review your class notes on homogeneous equations)
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  3. #3
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    I only have one example.
    and its my first week of Differential Equations.
    I just want to make sure i understand it before my next lesson which is tomorrow.

    Do i expand the brackets?
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  4. #4
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    Quote Originally Posted by wonderboy1953 View Post
    To guide you, why are you making that substitution? (review your class notes on homogeneous equations)
    I would bet he/she is making the substitution because

    Quote Originally Posted by BabyMilo View Post
    Use the substiution y=xu to find the general solution of the differential equation:
    BabyMilo,

    (x+xu)dy-xudx=0

    dy=udx+xdu

    (x+xu)(udx+xdu)-xudx=0\Rightarrow x(1+u)(udx+xdu)-xudx=0

    \Rightarrow (1+u)(udx+xdu)-udx=0

    Distribute and group us with du and xs with dx.

    Then integrate and back sub.
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  5. #5
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    not sure what Distribute and group us with du and xs with dx. means.

    should i come to u^2+(x+xu)(du/dx)=0 ?
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  6. #6
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    I am sorry I made a clerical error.

    \displaystyle (x+y)\frac{dy}{dx}=y\Rightarrow(x+y)dy=ydx

    y=ux \ \ \ dy=xdu+udx

    xdy+ydy=ydx\Rightarrow x(xdu+udx)+ux(xdu+udx)=uxdx

    x^2du+xudx+ux^2du+u^2xdx=uxdx

    x^2du+ux^2du+u^2xdx=0\Rightarrow (x^2+ux^2)du=-u^2xdx

    \displaystyle \frac{1+u}{u^2}du=-\frac{dx}{x}

    Sorry about the mistake earlier. Do you understand now?
    Last edited by dwsmith; Jan 20th 2011 at 01:33 PM. Reason: Fixed error
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