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Math Help - 2nd order differential equation

  1. #1
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    2nd order differential equation

    t^2*y'' - 2t*y' + y = 0

    To get the solution do you need to do a reduction of order? If so how do you get one solution, trial and error?
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  2. #2
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    Quote Originally Posted by Obstacle1 View Post
    t^2*y'' - 2t*y' + y = 0

    To get the solution do you need to do a reduction of order? If so how do you get one solution, trial and error?
    Well, if you happen to stumble over a solution reduction of order will work well. Here's a general method.

    Note that the order of the coefficient of y^{(n)} is n. Thus this is an Euler differential equation.

    So, make a change of variables:
    t = e^x \implies dt = e^x~dx \implies \frac{d}{dt} = e^{-x}\frac{d}{dx}

    And
    \frac{d^2}{dt^2} = \frac{d}{dt} \frac{d}{dt} = e^{-x}\frac{d}{dx} \left ( e^{-x} \frac{d}{dx} \right )

    \frac{d^2}{dt^2} = e^{-x} \left ( -e^{-x} \frac{d}{dx} + e^{-x} \frac{d^2}{dx^2} \right )

    \frac{d^2}{dt^2} = -e^{-2x} \frac{d}{dx} + e^{-2x} \frac{d^2}{dx^2}

    So the differential equation becomes:
    t^2*y^{\prime \prime} - 2t*y^{\prime} + y = 0

    \left ( e^{x} \right )^2 \cdot \left (-e^{-2x} \frac{dy}{dx} + e^{-2x} \frac{d^2y}{dx^2} \right ) - 2e^x \cdot e^{-x}\frac{dy}{dx} + y = 0

    \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + y = 0

    This is a linear homogeneous differential equation with constant coefficients and is easy to solve. So solve it and re-sub in x = ln(t).

    -Dan
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  3. #3
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    Quote Originally Posted by Obstacle1 View Post
    t^2*y'' - 2t*y' + y = 0

    To get the solution do you need to do a reduction of order? If so how do you get one solution, trial and error?
    The characheristic is:
    k(k-1)-2k+1=0\implies k^2-3k+1=0

    Thus, the solutions on \mathbb{R} - \{ 0 \} are given by:

    y = C_1|x|^{\frac{3+\sqrt{5}}{2}}+C_2 |x|^{\frac{3-\sqrt{5}}{2}}
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    The characheristic is:
    k(k-1)-2k+1=0\implies k^2-3k+1=0

    Thus, the solutions on \mathbb{R} - \{ 0 \} are given by:

    y = C_1|x|^{\frac{3+\sqrt{5}}{2}}+C_2 |x|^{\frac{3-\sqrt{5}}{2}}

    Did you derive the characteristic equation immediately? If so, how did you do it?
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