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Math Help - differential equations Q2

  1. #1
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    differential equations Q2

    losve the differnetial equation
    y' + 2y =2e^x
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  2. #2
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    Quote Originally Posted by twilightstr View Post
    is the answer y= 4/3e^4x + 2Ce^x
    I get \frac{2}{3}e^x + Ce^{-2x}

    Looking at the equation we have

    \frac{dy}{dx} + 2y = 2e^x

    So the integrating factor is e^{\int{2\,dx}} = e^{2x}.

    Multiplying both sides of the equation by the integrating factor gives

    e^{2x}\frac{dy}{dx} + 2e^{2x}y= 2e^{3x}

    The left hand side is a product rule expansion of e^{2x}y so we rewrite the equation as

    \frac{d}{dx}(e^{2x}y) = 2e^{3x}

    e^{2x}y = \int{2e^{3x}\,dx}

    e^{2x}y = \frac{2}{3}e^{3x} + C

    y = \frac{2}{3}e^x + Ce^{-2x}.
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  3. #3
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    i messed up on the algebra
    thanks
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