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Math Help - [SOLVED] Differentiation through Seperation

  1. #1
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    [SOLVED] Differentiation through Seperation

    Trying to solve y'=(e^X) (Y^3)

    I will post all my work so hopefully I can figure out what I'm doing wrong.

    dy/dx=e^xy^3
    y^-3 dy=∫ e^X dx
    y^-2/2=e^x+C1
    2/y^2=e^x+C
    1/y^2=(e^x+C)/2
    y^2=2/(e^x+C)
    Y=sqrt(2/(e^x+C)
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  2. #2
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    Hello, Iceman75x!

    Trying to solve: . \frac{dy}{dx}\:=\:e^xy^3

    I will post all my work so hopefully I can figure out what I'm doing wrong.

    \frac{dy}{dx}\:=\:e^xy^3

    y^{-3}\,dy\:=\:e^x\,dx

    \int y^{-3}dy \:=\:\int e^x\,dx

    \frac{y^{-2}}{2} \:=\:e^x + C_1 . . . . here ... and the next step, too

    It should have been: . \frac{y^{-2}}{{\color{red}-}2} \:=\:e^x + C_1 \quad\Rightarrow\quad y^{-2} \:=\:-2e^x + C

    . . \frac{1}{y^2} \:=\:C - 2e^x\quad\Rightarrow\quad y^2 \:=\:\frac{1}{C-2e^x} \quad\Rightarrow\quad\boxed{ y \;=\;\pm\frac{1}{\sqrt{C-2e^x}}}

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