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Math Help - Simple DE

  1. #1
    MHF Contributor alexmahone's Avatar
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    Simple DE

    (This problem is from the chapter Substitution methods and exact equations.)

    Find the general solution of (x+y)y'=x-y.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Simple DE

    Ok, let me give it a shot.

    Let x+y=v.

    1+\frac{dy}{dx}=\frac{dv}{dx}

    \frac{dy}{dx}=\frac{dv}{dx}-1

    Substituting, we get

    v\left(\frac{dv}{dx}-1\right)=x-(v-x)=2x-v

    v\frac{dv}{dx}=2x

    \int vdv=\int 2xdx

    \frac{v^2}{2}=x^2+K

    \frac{(x+y)^2}{2}=x^2+K

    x^2+2xy+y^2=2x^2+2K

    x^2-2xy-y^2=-2K

    x^2-2xy-y^2=C, where C=-2K.
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  3. #3
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    Re: Simple DE

    Quote Originally Posted by alexmahone View Post
    Ok, let me give it a shot.

    Let x+y=v.

    1+\frac{dy}{dx}=\frac{dv}{dx}

    \frac{dy}{dx}=\frac{dv}{dx}-1

    Substituting, we get

    v\left(\frac{dv}{dx}-1\right)=x-(v-x)=2x-v

    v\frac{dv}{dx}=2x

    \int vdv=\int 2xdx

    \frac{v^2}{2}=x^2+K

    \frac{(x+y)^2}{2}=x^2+K

    x^2+2xy+y^2=2x^2+2K

    x^2-2xy-y^2=-2K

    x^2-2xy-y^2=C, where C=-2K.
    Hi alexmahone,

    Your method is correct. Alternatively you could use the substitution, u=\frac{y}{x} as this is a differential equation of the "Homogeneous type".
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