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Math Help - DE Initial Value Problem

  1. #1
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    DE Initial Value Problem

    Question: dy/dx= e^2x+y , y(0)=3
    using variable seperable
    i have solved till ln y = 1/2e^2x+c then i am stuck,would any kind soul help me to further solve it?
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  2. #2
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    Is your equation \displaystyle \frac{dy}{dx}= e^{2x}+y

    or \displaystyle \frac{dy}{dx}= e^{2x+y} ??
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  3. #3
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    oh it the first 1
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  4. #4
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    \displaystyle \frac{dy}{dx} = e^{2x} + y

    \displaystyle \frac{dy}{dx} - y = e^{2x}.

    This is first order linear, so multiplying both sides by the integrating factor \displaystyle e^{\int{-1\,dx}} = e^{-x} gives

    \displaystyle e^{-x}\,\frac{dy}{dx} - e^{-x}\,y = e^{x}

    \displaystyle \frac{d}{dx}\left(e^{-x}\,y\right) = e^x

    \displaystyle e^{-x}\,y = \int{e^x\,dx}

    \displaystyle e^{-x}\,y = e^x + C

    \displaystyle y = e^{2x} + Ce^x.


    Now, you know from your initial condition that when \displaystyle x = 0, y = 3, so

    \displaystyle 3 = e^{2\cdot 0} + Ce^0

    \displaystyle 3 = 1 + C

    \displaystyle C = 2.


    Therefore the solution is

    \displaystyle y = e^{2x} + 2e^{x}.
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  5. #5
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    i see thanks
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