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Math Help - Find the particular solution of the differential equation

  1. #1
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    Find the particular solution of the differential equation

    Hi again,

    \displaystyle y(2)=2
    \displaystyle y'+\frac{1}{x}y=0
    \displaystyle P(x)=\frac{1}{x},Q(x)=0,U(x)=e^{\int\frac{1}{x}dx}
    \displaystyle y=\frac{1}{U(x)}\int Q(x)U(x)dx
    \displaystyle y=\frac{1}{x}\int 0 x dx
    \displaystyle y=\frac{1}{x}\int dx

    Is this the way to go about the problem?
    I think this is the set up by first-order linear differential equation. Im having problems with finding the solution.
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  2. #2
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    You have found that your integrating factor is \displaystyle e^{\int{\frac{1}{x}\,dx}} = e^{\ln{x}} = x.

    So multiplying through by the integrating factor gives

    \displaystyle x\,\frac{dy}{dx} + y = 0

    \displaystyle \frac{d}{dx}(x\,y) = 0

    \displaystyle x\,y = \int{0\,dx}

    \displaystyle x\,y = C

    \displaystyle y = \frac{C}{x}.


    Now substitute your boundary condition to find \displaystyle C.
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