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Math Help - Equations

  1. #1
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    Equations

    Hello, was just wondering if someone could give me some help.
    i have to find the solution of the equation
    y'-2xy=x which satisfies y(0)=0

    so far i have done the basic and written y'= 2xy + x
    next i was going to divide all through by y but got a bit stuck after that point. it probably is a relatively easy question but i just am not sure where to go next... if anyone could lend a hand it would be appreciated!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by biffer View Post
    Hello, was just wondering if someone could give me some help.
    i have to find the solution of the equation
    y'-2xy=x which satisfies y(0)=0

    so far i have done the basic and written y'= 2xy + x
    next i was going to divide all through by y but got a bit stuck after that point. it probably is a relatively easy question but i just am not sure where to go next... if anyone could lend a hand it would be appreciated!
    Have you tried an integrating factor?

    -Dan
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  3. #3
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    thanks!
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  4. #4
    Super Member wingless's Avatar
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    You can also try solving the equation without integrating factor..

    y' - 2 x y = x

    First solve the LHS..

    y' - 2 x y = 0

    y' = 2 x y

    \frac{dy}{y} = 2x~dx

    \int \frac{dy}{y} = \int 2x~dx

    ln|y|=x^2+C

    y=C\cdot e^{x^2}.... \boxed{I}

    Now take C as a function, and find y'

    y' = (C')(e^{x^2}) + (C)(2xe^x)

    y' = C'e^{x^2} + 2Cxe^x.... \boxed{II}

    Put y and y' (marked as I and II) in the original equation:

    y' - 2 x y = x

    C'e^{x^2} + 2Cxe^x - 2 x C e^{x^2} = x

    C'e^{x^2} = x

    dC = \frac{x}{e^{x^2}}~dx

    C = -\frac{1}{2e^{-x^2}} + K

    Now put this C in I,

    y=Ce^{x^2}

    y=(-\frac{1}{2e^{-x^2}} + K)e^{x^2}

    y= -\frac{1}{2} + K\cdot e^{x^2}
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