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Thread: 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 $\displaystyle (x+y)y'=x-y$.
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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 $\displaystyle x+y=v$.

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

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

    Substituting, we get

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

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

    $\displaystyle \int vdv=\int 2xdx$

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

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

    $\displaystyle x^2+2xy+y^2=2x^2+2K$

    $\displaystyle x^2-2xy-y^2=-2K$

    $\displaystyle x^2-2xy-y^2=C$, where $\displaystyle C=-2K$.
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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 $\displaystyle x+y=v$.

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

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

    Substituting, we get

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

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

    $\displaystyle \int vdv=\int 2xdx$

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

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

    $\displaystyle x^2+2xy+y^2=2x^2+2K$

    $\displaystyle x^2-2xy-y^2=-2K$

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

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