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Math Help - Differential equation on -x^2

  1. #1
    Newbie Trikotnik's Avatar
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    Differential equation on -x^2

    The differential equation y' + 2xy = 4e on -x^2 is given.

    Solve the modified homogeneous equation.
    What is the general solution to the equation?
    Find that particulate solution which goes through T(0,2).
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Trikotnik View Post
    The differential equation y' + 2xy = 4e on -x^2 is given.

    Solve the modified homogeneous equation.
    What is the general solution to the equation?
    Find that particulate solution which goes through T(0,2).
    What does "on -x^2" mean?

    CB
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  3. #3
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    Question

    First, I assume you ment y'+2xy=4e^{-x^2} .

    Anyway, the equation is a first order linear differential equation. In general it can be written down as y' + p(x)y=q(x). Using the variation of the constant method general solution formula can be found, and it goes something like this y=\left(\int q(x)\cdot e^{-P(x)} dx+C\right)\cdot e^{P(x)} where P is the primitive function of the function p, \qquad \int p(x) dx= P(x)+const..

    Using the formula should suffice to solve the problem(s).
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Trikotnik View Post
    The differential equation y' + 2xy = 4e on -x^2 is given.

    Solve the modified homogeneous equation.
    What is the general solution to the equation?
    Find that particulate solution which goes through T(0,2).
    If as MathoMan suggests you mean:

    y'+2xy=e^{-x^2}

    we can use an integrating factor of e^{x^2} to get:

    \dfrac{d}{dx}\left( e^{x^2}y(x)\right)=1

    so:

    y(x)=e^{-x^2}(x+c)

    CB
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