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Thread: Trouble with the second root

  1. #1
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    Trouble with the second root

    Original equation: xy'' + y' + 0. I have found that the roots are both 0, and one of the solutions is:

    $\displaystyle y_1=\sum_{m=0}^\infty \frac{(-1)^m a_0}{(m!)^2}x^m$

    That solution checks. For y2, I tried using $\displaystyle y_2 = y_1 \int \frac{e^{-\int P(x) dx}}{{y_1}^2}dx$, but using P(x) = 1/x, y2 turned into:

    $\displaystyle y_2 = y_1(ln(x) + \frac{x}{2} + \frac{19x^2}{288} + \frac{7x^3}{864} + ...)$

    which is not even close to the answer in the back of the book. Please advise.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Trouble with the second root

    If the equation is:

    $\displaystyle x\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$

    then I would simply, rewrite the left side as:

    $\displaystyle \frac{d}{dx}\left(x\frac{dy}{dx}\right)=0$

    $\displaystyle \int\,d\left(x\frac{dy}{dx}\right)=\int 0\,dx$

    $\displaystyle x\frac{dy}{dx}=c_1$

    separate variables, losing a trivial solution:

    $\displaystyle dy=\frac{c_1}{x}\,dx$

    $\displaystyle \int\,dy=c_1\int\frac{1}{x}\,dx$

    $\displaystyle y=c_1\ln|x|+c_2$

    Obviously, though, I am missing something...
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  3. #3
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    Re: Trouble with the second root

    I mistyped the OP. The differential equation is xy'' + y' + y = 0. (Otherwise, I just set w = y' and solve.)

    Everything else is written as intended.
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  4. #4
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Trouble with the second root

    Quote Originally Posted by phys251 View Post
    The differential equation is xy'' + y' + y = 0.
    $\displaystyle y(x)= c_1 J_0\left(2 \sqrt{x}\right)+2 c_2 Y_0\left(2 \sqrt{x}\right)$
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