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Math Help - find particular solution for first order linear DE

  1. #1
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    Exclamation find particular solution for first order linear DE

    (1+x)y' + y = cos(x).... y(0) = 1

    So far, I have:
    P(x) = 1/(1+x)
    Q(x) = cos(x)/(1+x)
    p(x) = e^int(P(x))dx = e^int(1/1+x)dx = e^(ln(1+x))
    ... I think I did the integral wrong. It's been half a year, my integration skills need to worked on. But this homework is due in an hour and if somebody could real quickly guide me thru it and then I can work thru this section on my own when I don't have pressure to turn homework in... I would really appreciate it!

    Once I have p(x), I multiply both sides of DE by it.
    Then I'm supposed to "recognize" that left side is a derivative, so taking integral of both sides just means taking integral of right side.

    Once I do that, I'll have a C, so is that when I plug in the initial condition?
    And then once I get C and substitue that number in, that's my particular solution?

    Thanks!
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  2. #2
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    You have gotten the DE to the point where

    \frac{dy}{dx} + \left(\frac{1}{1 + x}\right)y = \frac{\cos{x}}{1 + x}

    and so the integrating factor is e^{\int{\frac{1}{1 + x}\,dx}} = e^{\ln{(1 + x)}} = 1 + x.


    Multiplying through by the integrating factor gives

    (1 + x)\frac{dy}{dx} + y = \cos{x}

    \frac{d}{dx}[(1 + x)y] = \cos{x}

    (1 + x)y = \int{\cos{x}\,dx}

    (1 + x)y = \sin{x} + C

    y = \frac{\sin{x} + C}{1 + x}.


    Plugging in the initial condition y(0) = 1 gives

    1 = \frac{\sin{0} + C}{1 + 0}

    1 = \frac{0 + C}{1}

    1 = C.


    Therefore y = \frac{\sin{x} + 1}{1 + x}.
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