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Math Help - differential help

  1. #1
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    differential help

    solve when x=0 and y= 2




    ok I have setup as shown`(is this ok?). is there some rule about F/F'? to make it easier to solve?

    many thanks
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  2. #2
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    Re: differential help

    Quote Originally Posted by decoy808 View Post
    solve when x=0 and y= 2




    ok I have setup as shown`(is this ok?). is there some rule about F/F'? to make it easier to solve?

    many thanks
    \frac{2y+4}{y^2+4y}dy=dx

    Integrating both sides,

    ln(y^2+4y)=x+C

    Substitute x=0 and y=2 to find C.
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  3. #3
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    Re: differential help

    \displaystyle \begin{align*} (2y + 4)\,\frac{dy}{dx} &= y^2 + 4y \\ \left[\frac{2y + 4}{y(y + 4)}\right]\,\frac{dy}{dx} &= 1 \\ \int{\left[\frac{2y + 4}{y(y+4)}\right]\,\frac{dy}{dx}\,dx} &= \int{1\,dx} \\ \int{\frac{2y + 4}{y(y + 4)}\,dy} &= \int{1\,dx}\end{align*}

    Go from here, you will need to use a substitution or Partial Fractions on the LHS.
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  4. #4
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    Re: differential help

    Quote Originally Posted by decoy808 View Post
    solve when x=0 and y= 2




    ok I have setup as shown`(is this ok?). is there some rule about F/F'? to make it easier to solve?

    many thanks
    No, there is no rule about F/F'. There is, however, a rule about F'/F which this is. If you let u= F, then du= F' dx so
    \int \frac{F'}{F}dx= \int \frac{1}{u}du= ln|u|+ C= ln|F(x)|+ C.
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