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Math Help - differential equation need Help

  1. #1
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    differential equation need Help

    dy/dx = -6xy
    \int \frac{1}{y} dy = \int -6x dx
    ln y = -3x^2 + c
    y = e^{-3x^2 + c}

    is my answer Correct ?
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  2. #2
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    Re: differential equation need Help

    Yes, though it can be taken one step further,

    y = e^{-3x^{2}+c}=e^{-3x^{2}}.e^{c}=Ae^{-3x^{2},

    which, subsequently, will be slightly easier to deal with.
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  3. #3
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    Re: differential equation need Help

    its given y(0) = 7 ,to find value of C

    7 = e^{-3x^2} \times e^c

    Then ?
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  4. #4
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    Re: differential equation need Help

    Quote Originally Posted by mastermin346 View Post
    its given y(0) = 7 ,to find value of C

    7 = e^{-3x^2} \times e^c

    Then ?
    y(0) = 7 says that when x = 0 y = 7. What change does that make in the line you wrote?

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  5. #5
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    Re: differential equation need Help

    7 = e^{-3(0)^2} \times e^c
    7 = 1 \times e^c
    C = ln 7

    right or wrong ?
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  6. #6
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    Re: differential equation need Help

    y=Aexp (-3x^2)
    y (0)=7, so 7=A.
    y=7exp (-3x^2).
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  7. #7
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    Re: differential equation need Help

    Quote Originally Posted by mastermin346 View Post
    dy/dx = -6xy
    \int \frac{1}{y} dy = \int -6x dx
    ln y = -3x^2 + c You are wrong in this step
    y = e^{-3x^2 + c}

    is my answer Correct ? You need to go further
    First note that \displaystyle \begin{align*} \int{\frac{1}{y}\,dy} = \ln{|y|} \end{align*}, NOT \displaystyle \begin{align*} \ln{(y)} \end{align*} so this means the correct working out would be

    \displaystyle \begin{align*} \ln{|y|} &= -3x^2 + C \\ |y| &= e^{-3x^2 + C} \\ |y| &= e^{C}\,e^{-3x^2} \\ y &= \pm e^{C}\,e^{-3x^2} \\ y &= A\,e^{-3x^2} \textrm{ where } A = \pm e^C \end{align*}
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