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Math Help - First Order Separable Differential equation

  1. #1
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    Question First Order Separable Differential equation

    Hello,
    I'm in Differential Equations and I have my midterm tomorrow... I can do most everything on my review sheet, but I'm a bit confused when it comes to a very simple initial value problem:

    y' = x*y^3, y(0)=1

    Here is what I'm doing:
    I figure I can use separation of variables like so:

    dy/y^3 = x*dx

    ..and after integrating both sides:

    -1/(2*y^2) = (x^2)/2

    solving for y gives:

    y = 1/x

    ...which the initial condition y(0) = 1 does not hold.

    Can somebody point out what I'm doing wrong? I'm quite confused and I'm running out of time...

    Thanks in advance,
    - Jeff
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  2. #2
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    Lexington, MA (USA)
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    Hello, ibanez270dx!

    You forgot the "plus C" . . .


    y' \:=\: xy^3,\quad y(0)\,=\,1

    Here is what I'm doing: . \frac{dy}{y^3} \:=\: x\,dx

    Integrating: . -\frac{1}{2y^2} \:= \:\frac{x^2}{2} \:{\color{red}+ \;C}

    Multiply by -2: . \frac{1}{y^2} \:=\:-x^2 + C

    Invert: . y^2 \:=\:\frac{1}{C - x^2}


    Now plug in the initial value . . .

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  3. #3
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    thanks! wow, that totally makes a lot more sense. I knew it had to be something simple.

    Thanks again, your help is much appreciated!!
    - Jeff
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