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Math Help - FP1 first order differential help please

  1. #1
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    FP1 first order differential help please

    find y in terms of x given that x(dy/dx)+3y=e^x

    And that when y=1 x=1
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  2. #2
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    Quote Originally Posted by Bryn View Post
    find y in terms of x given that x(dy/dx)+3y=e^x

    And that when y=1 x=1
    x\frac{dy}{dx}+3y=e^x

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

    Now multiply by the integrating factor e^{\int \frac3{x} \, dx} = x^3.

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

    \int d(yx^3)=\int x^2 e^x \, dx

    yx^3=(x^2 - 2x + 2)e^x + C
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  3. #3
    Junior Member rednest's Avatar
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    Smile I have this exam next monday too.

    Quote Originally Posted by Bryn View Post
    find y in terms of x given that x(dy/dx)+3y=e^x

    And that when y=1 x=1
    x\frac{dy}{dx} + 3y = e^x
    \frac{dy}{dx} + \frac{3}{x}y = \frac{e^x}{x}
    Integrating factor = e^{\int \frac{3}{x}\, dx}=e^{ln x^3} = x^3
    Multiply through x^3, giving x^3\frac{dy}{dx} + 3x^2y = e^xx^2
    So x^3 y = \int x^2e^x\, dx

    I would \int x^2 e^x \, dx by parts:
    u=x^2, \frac{du}{dx} = 2x, \frac{dv}{dx} = e^x, v = e^x

    so x^3y = e^xx^2 - \int 2xe^x\, dx
    u=2x, \frac{du}{dx}=2, \frac{dv}{dx}=e^x, v=e^x
    x^3y = e^xx^2 - 2xe^x + \int 2e^x \,dx = (x^2-2x+2)e^x + C

    To get C, substitute y=1, x=1 into x^3y = (x^2-2x+2)e^x + C

    So 1=(1-2+2)e + C => C = 1-e

    x^3y=(x^2-2x+2)e^x + 1-e
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