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Math Help - I need some help with 3 prob.s from my AP class...

  1. #1
    Newbie woohoo's Avatar
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    Question I need some help with 3 prob.s from my AP class...

    Solve the following using separation of variables. Let C represent an arbitrary constant.

    1) (11 + x^2)y ' = 11x^(3)*y



    ^ I got (Ce^((11x^2)/2))/(e^(11log(x^2+11))) but that's not correct.


    Solve the initial value problem.

    2) y ' - 2y + 4 = 0, y(1) = 12

    ^I don't even know how to approach this one.



    3)y^2 * dy/dx = x^(-3), y(5) = 0



    ^I got ((25-x)/6)^(1/3) but that's wrong.




    I'm not necessarily looking for the answers...but take me step my step? Or atleast guide me through one or two of them?

    Thanks ahead of time!
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  2. #2
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    Hello, woohoo!

    1)\;\;(11 + x^2)\,\frac{dy}{dx}\:=\: 11x^3y

    Separate variables: . \frac{dy}{y} \;=\;11\,\frac{x^3}{x^2+11} \quad\Rightarrow\quad \frac{dy}{y} \;=\;11\left(x - \frac{11x}{x^2+11}\right)

    Integrate: . \int\frac{dy}{y} \;=\;11\left[\int x\,dx - 11\int\frac{x\,dx}{x^2+11}\right]


    Therefore: . \ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C




    2)\;\;\frac{dy}{dx} - 2y + 4 \:=\: 0,\quad y(1) = 12
    We have: . \frac{dy}{dx} - 2y \:=\:-4

    Integrating factor: . I \:=\:e^{\int(-2)\,dx} \:=\:e^{-2x}

    Multiply by I\!:\;\;e^{-2x}\,\frac{dy}{dx} - 2e^{-2x}y \;=\;-4e^{-2x} \quad\Rightarrow\quad \frac{d}{dx}\left(e^{-2x}y\right) \;=\;-4e^{-2x}

    Integrate: . e^{-2x}y \;=\;2e^{-2x} + C \quad\Rightarrow\quad y \;=\;2 + Ce^{2x}

    Since y(1) = 12, we have: . 12 \;=\;2 + Ce^2 \quad\Rightarrow\quad C \:=\:\frac{10}{e^2}

    Therefore: . y \;=\;2 + \frac{10}{e^2}\,e^{2x} \quad\Rightarrow\quad y \;=\;2 + 10e^{2x-2}




    3)\;\;y^2\,\frac{dy}{dx} \:=\:x^{-3},\quad y(5) = 0

    Separate variables: . y^2dy \;=\;x^{-3}dx

    Integrate: . \frac{y^3}{3} \;=\;\frac{x^{-2}}{-2} + C \quad\Rightarrow\quad y^3 \;=\;-\frac{3}{2x^2} +C

    Since y(5) = 0, we have: . 0^3 \;=\;-\frac{3}{2(5^2)} + C \quad\Rightarrow\quad C \:=\:\frac{3}{50}

    Therefore: . y^3 \;=\;-\frac{3}{2x^2} + \frac{3}{50}

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  3. #3
    Newbie woohoo's Avatar
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    Sweet!

    Now, if i were to multiply both sides of the first one by e, I would get...


    y=Ce^(11x^2)/e^(11ln(x^2+11)/2) correct?

    how can I simplify this?

    would it be:

    2Ce^((11x^2)/2)/(11x^2+121) ?
    Last edited by woohoo; December 4th 2008 at 09:55 PM.
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