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Math Help - Find the solution that solves the initial condition.

  1. #1
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    Find the solution that solves the initial condition.

    Differential Equation: 2xy' - ln (x^2) = 0, y(1)=8 Initial Condition
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  2. #2
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    Quote Originally Posted by gingerbailey View Post
    Differential Equation: 2xy' - ln (x^2) = 0, y(1)=8 Initial Condition
    The DE can be written as \frac{dy}{dx} = \frac{\ln x^2}{2x} = \frac{\ln |x| }{x} \Rightarrow y = \int \frac{\ln |x|}{x} \, dx.

    To solve the integral make the substitution u = \ln |x| \Rightarrow \frac{du}{dx} = \frac{1}{|x|}.

    Note that |x| = x when x > 0 and |x| = -x when x < 0.
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    Well, we have:

    2xy' - \ln{(x^2)} = 0

    This is pretty simple to organize, using basic algebra you get:

    2xy' = \ln{(x^2)}

    y' = \frac{\ln{(x^2)}}{2x}

    Now, we integrate both sides:

    \int y' ~dx = \int \frac{\ln{(x^2)}}{2x} ~dx

    Integrating, we get:

    y = \frac{\ln^2{(x^2)}}{8} + C

    We want to find the solution for when y(1) = 0, so we plug and solve:

    y(1) = \frac{\ln^2{(1^2)}}{8} + C = 8

    = \frac{0}{8} + C = 8

    C = 8

    So, our solution is:

    y = \frac{\ln^2{(x^2)}}{8} + 8

    And there you go.
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