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Thread: 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 $\displaystyle \displaystyle \frac{dy}{dx}= e^{2x}+y$

    or $\displaystyle \displaystyle \frac{dy}{dx}= e^{2x+y}$ ??
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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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    $\displaystyle \displaystyle \frac{dy}{dx} = e^{2x} + y$

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

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

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

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

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

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

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


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

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

    $\displaystyle \displaystyle 3 = 1 + C$

    $\displaystyle \displaystyle C = 2$.


    Therefore the solution is

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