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2nd ode proof.

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data
    suppose [itex]v(t) , u(t) [/itex]are two linearly independent solution of the 2nd DE.
    [itex](6t^2-t-1)y''+t^2e^ty'-(3t^3-t-1)y=t^2e^t-3t^3+1[/itex]
    satisfying the condition [itex]v(0)=u(0)=1 [/itex], prove that[itex] u'(0) ≠ v'(0)[/itex]


    2. Relevant equations



    3. The attempt at a solution
    I've tried to use Wronskian, but it seems to fail.
     
  2. jcsd
  3. Oct 19, 2012 #2

    Simon Bridge

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    Welcome to PF;
    If u and v are solutions to the non-homogeneous DE, then u-v is...
     
  4. Oct 19, 2012 #3
    You need to use a theorem on when solutions to an ODE are unique.
     
  5. Oct 19, 2012 #4
    thank you Simon, yep , u-v should be one of the general solution of the corresponding homogeneous equation. But how can I find the other one ,?

    deluks ,what do you mean ? (or what theorem?

    In this question,let the particular solution be x(t), then
    u-x, v-x is the general solution of the corresponding homogeneous equation . Are they linearly independent ? (for general case, Are they always linearly independent ?)
     
  6. Oct 19, 2012 #5

    Simon Bridge

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    I was thinking: if ##y_1## and ##y_2## are solutions to the corresponding homogeneous equations, then surely ##u-v=c_1y_1+c_2y_2## ... what happens if you differentiate both sides? What has to happen to make u'=v'?

    This approach may not be rigorous enough for your course...

    You really need to show that, according to some method for demonstrating that two solutions are independent, that they cannot have the same derivative. i.e. If u and v are independent, then they satisfy some criteria ... if their derivatives are the same they cannot satisfy those criteria.

    You know what it takes to make two solutions independent right?
    You must know more than one way of testing independence?
     
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