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Solution -> Second Order ODE?

  1. Mar 28, 2008 #1
    Hi, I am trying to decide whether y(t) = c1t^2 + c2 t^−1, where c1 and c2 are arbitrary constants, is the general solution of the differential equation (t^2)y'' − 2y = 0 for t > 0 and justify the answer, but I don't really know how to approach it from this "side" of the problem.

    Any suggestions would be greatly appreciated.
     
    Last edited: Mar 28, 2008
  2. jcsd
  3. Mar 28, 2008 #2

    Hootenanny

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    To check that the proposed solution is a solution of the ODE, one only needs to substitute the proposed solution into the ODE and verify that the equation holds.
     
  4. Mar 28, 2008 #3
    Like Hootenanny said, i am just going to give more details:[tex]y(t)=c_1t^{2}+c_2t^{-1}[/tex]

    [tex]y'(t)=2c_1t-c_2t^{-2},\ \ \ y''(t)=2c_1+2c_2t^{-3}[/tex]

    [tex]t^{2}y''-2y=0[/tex] Now all you need to do is plug in for y'', and y. That is

    [tex]t^{2}(2c_1+2c_2t^{-3}-2(c_1t^{2}+c_2t^{-1})=.....[/tex]

    NOw if this equals 0, then we conclude that :[tex]y(t)=c_1t^{2}+c_2t^{-1}[/tex]
    is a solution of

    [tex]t^{2}y''-2y=0[/tex]
     
  5. Mar 28, 2008 #4

    HallsofIvy

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    sutupidmath showed that that is a solution to the differenital equation. The fact that it is the general solution follows from the fact that this is a second order linear differential equation- so its solution set is a two dimensional vector space- and the two given functions are independent- and so form a basis for the solution space.
     
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