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Math Help - Solve y'=y^2 subject to y(0)=yo.

  1. #1
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    Solve y'=y^2 subject to y(0)=yo.

    So I need to solve y'=y^2 subject to y(0)=yo (ynaught)

    So I know I can separate and get y=1/(C-x) or y=1/((1/yo)-x) after applying the condition.

    My question is, why can't I just integrate to get y=y^3/3+C? Why do I have to separate?


    The next question is to solve A'(t)=6sqrt(A)(9-A) with A(0)=1, and I can't figure out how I would separate this one because I'd get dA/(6sqrt(A)(9-A))=dt, but the LHS is impossible to integrate simply. So can I just take integral 6sqrt(A)(9-A) to do this?
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  2. #2
    Junior Member RaisinBread's Avatar
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    Well, it is true that \int y^2 dy = \frac {\ y^3}{3} +C, but that is not the answer to the first question you mention because the above integral is not what you are being asked to solve.

    In y' = y^2, y' is in fact  \frac {\ dy}{dx}, so you have to move y^2 with dy, and get;
    \frac {\ dy}{y^2}=dx, and then, integrate on both sides : \int \frac {\ dy}{y^2}= \int dx

    Trying to "simply integrate" as you suggest would amount to attempt to the following:
    \int \frac {\ dy}{dx} = \int y^2
    These two expressions are not defined.
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    Ok, I got that now thanks.

    The thing is for A'(t)=6sqrt(A)(9-A) with A(0)=1 wouldn't I also have to separate. That gives dA/(6sqrt(A)(9-A))=dt. This is my primary concern...How would the LHS be possible to integrate. Someone told me I can just integrate 6sqrt(A)(9-A), but that wouldn't be separating variables...
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by twittytwitter View Post
    My question is, why can't I just integrate to get y=y^3/3+C? Why do I have to separate?

    Trying to "simply integrate" as you suggest would amount to attempt to the following:
    \int \frac {\ dy}{dx} = \int y^2
    These two expressions are not defined.
    I think there may be a subtle point missed in here, so I'll add one more bit.

    twittytwitter you want to integrate y' = y^2, the left side becoming (1/3)y^3 + C, but you aren't integrating over y...

    \displaystyle \frac{dy}{dx} = y^2

    \displaystyle \int \frac{dy}{dx}~dx = \int y^2~dx

    Notice that we are integrating over x, not y. The LHS is now \displaystyle \int \frac{dy}{dx}~dx = y + C, but the RHS is a mess: \int y^2~dx

    -Dan
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  5. #5
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    Quote Originally Posted by twittytwitter View Post
    Ok, I got that now thanks.

    The thing is for A'(t)=6sqrt(A)(9-A) with A(0)=1 wouldn't I also have to separate. That gives dA/(6sqrt(A)(9-A))=dt. This is my primary concern...How would the LHS be possible to integrate. Someone told me I can just integrate 6sqrt(A)(9-A), but that wouldn't be separating variables...
    \dfrac{dA}{\sqrt{A}(9-A)} = 6 \, dt

    left side ... let u = \sqrt{A}

    du = \dfrac{dA}{2\sqrt{A}}

    dA = 2u \, du

    substitute ...

    \dfrac{2u}{u(9-u^2)} \, du = 6 \, dt

    \dfrac{2}{9-u^2} \, du = 6 \, dt

    \dfrac{2}{(3-u)(3+u)} \, du = 6 \, dt

    use the method of partial fractions to integrate the left side
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