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Math Help - 2nd Order Diff. Equ. Solution

  1. #1
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    2nd Order Diff. Equ. Solution

    My goal is to find the general solution of

    x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

    What we've learned about 2nd order equations is this:

    1. How to reduce the order when y or x is absent from the equation
    2. How to solve equations in the form y'' + py' + qy = 0, where p and q are constants
    3. How to solve equations in the form y'' + py' +qy = ke^(ax), k*sin(b*x) + r*cos(b*x), or a + b*x + c*x^2, where a, b, c, k, and r are constants.
    4. How to find a second solution to the equation y'' + p(x)*y' + q(x)*y = 0, where p(x) and q(x) are functions of x, when one solution is already known.
    5. How to find a particular solution to the equation y'' + p(x)*y' + q(x)*y = r(x), when the general solution to y'' + p(x)*y' + q(x)*y = 0 is known.

    The problem is, I don't know how to reduce

    x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

    to an equation in any of the forms above. That is, I don't even know how to find the homogeneous equation, because I can't separate x out of the right side entirely.

    At this level of understanding, should I be using intuition to find one solution, or is there a way to use substitution, or what?

    It looked like I could factor the left side, up until the 2*y part.

    Anyway, I'm stumped. Any help is much appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by primasapere View Post
    My goal is to find the general solution of

    x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

    What we've learned about 2nd order equations is this:

    1. How to reduce the order when y or x is absent from the equation
    2. How to solve equations in the form y'' + py' + qy = 0, where p and q are constants
    3. How to solve equations in the form y'' + py' +qy = ke^(ax), k*sin(b*x) + r*cos(b*x), or a + b*x + c*x^2, where a, b, c, k, and r are constants.
    4. How to find a second solution to the equation y'' + p(x)*y' + q(x)*y = 0, where p(x) and q(x) are functions of x, when one solution is already known.
    5. How to find a particular solution to the equation y'' + p(x)*y' + q(x)*y = r(x), when the general solution to y'' + p(x)*y' + q(x)*y = 0 is known.

    The problem is, I don't know how to reduce

    x^2*y'' - 2*x*y' + 2*y = x*e^(-x)

    to an equation in any of the forms above. That is, I don't even know how to find the homogeneous equation, because I can't separate x out of the right side entirely.

    At this level of understanding, should I be using intuition to find one solution, or is there a way to use substitution, or what?

    It looked like I could factor the left side, up until the 2*y part.

    Anyway, I'm stumped. Any help is much appreciated.
    This is Cauchy's ODE, and you make the change of variable:

    t=ln(x),

    or equivalently:

    x=exp(t).

    Which transforms the quation in to a 2nd order ODE with constant
    coefficients. (see here)

    RonL
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  3. #3
    Global Moderator

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    Quote Originally Posted by primasapere View Post
    My goal is to find the general solution of

    x^2*y'' - 2*x*y' + 2*y = x*e^(-x)
    Actually it is simpler than CaptainBlank said.
    It is the Euler equation.

    Solve, the homogenous equation,
    k(k-1)-2k+2=0
    Thus,
    k^2-3k+2=0
    k=[3+/-sqrt(5)]/2

    Thus, the homogenous solution is,
    y=c_1 x^(3/2) cos(\sqrt(5)/2 ln |x|)+c_2 x^(3/2)sin(\sqrt(5)/2 \ln|x|)

    Now for the particular solution you can pretend it is of the form
    y=Axe^{-x}

    And take it from there.

    If not use Largrange's varaition of parameters techiqnue
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