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Math Help - Separable Equations

  1. #1
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    Separable Equations

    Solve the equation
    x dy/dx = y^2 - y
    1/x dx = 1/(y^2 -y) dy
    ln x + C = ln (y^2-y)
    e^l^n^x * e^C = y^2 - y
    Cx = y^2 - y
    y^2 - y -Cx = 0
    I'm pretty sure i did something wrong with this one


    Solve the equation
    e^xy dy/dx = e^-^y + e^-^2^x^-^y
    No idea where to start on this so any help is greatly appreciated
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  2. #2
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    pickslides's Avatar
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    I'm not really sure what you did, after separation you should have \displaystyle \frac{dy}{y^2-y}= \frac{dx}{x}
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Naples View Post
    Solve the equation
    x dy/dx = y^2 - y
    1/x dy/dx = 1/(y^2 -y) dy
    ln x + C = ln (y^2-y)
    e^l^n^x * e^C = y^2 - y
    Cx = y^2 - y
    y^2 - y -Cx = 0
    I'm pretty sure i did something wrong with this one


    Solve the equation
    \displaystyle e^xy dy/dx = e^-^y + e^-^2^x^-^y
    No idea where to start on this so any help is greatly appreciated
    \displaystyle  x\frac{dy}{dx}=y^2-y \iff \frac{dy}{y(y-1)}=\frac{dx}{x}

    Now by partial fractions or other algebraic manipulations we get

    \displaystyle  \left(\frac{-1}{y}+\frac{1}{y-1}\right)dy=\frac{dx}{x}

    Now we can integrate both sides

    For the 2nd one note that

    \displaystyle  e^{-y}+e^{-2x-y}=e^{-y}(1+e^{-2x})

    So once again you can separate the equation.
    Can you finish from here?
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  4. #4
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    e^xy dy/dx = e^-^y(1+ e^-^2^x)
    ye^ydy = (e^-x + e^-^3^x)dx
    using integration by parts
    ye^y - e^y = -e^-^x - (1/3)e^-^3^x + C
    e^y(y -1) = -e^-^x - (1/3)e^-^3^x + C
    Hopefully I did this right, if so would it be necessary to simplify this further?
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