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Second order differential equation

  1. Sep 19, 2007 #1
    Solve by direct substitution taht the function [tex]\phi[/tex](t) = Asin(wt) + Bcos(wt) is a solution of the second order differential equation [tex]\ddot{\phi}[/tex] = -w[tex]^{2}[/tex][tex]\phi[/tex]. ( Since this solution involves two arbitrary constants - the coeffecients of the sine and consine functions - it is in fact the general solution).

    I'm not sure what is meant by direct substitution..
    Can someone help me get started?
    Thanks alot!
     
  2. jcsd
  3. Sep 19, 2007 #2

    arildno

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    What does the word "substitution" normally mean?
     
  4. Sep 19, 2007 #3
    Is direct substitution different then normal?
    I tried subbing in Asin(wt) + Bcos(wt) into the second order differential equation but that got me nowhere...
     
  5. Sep 19, 2007 #4

    arildno

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    Well, you are asked to show that the second derivative of the given function equals the negative square of w times the given function.

    Have you tried to show that?
     
  6. Sep 19, 2007 #5
    That's why I'm confsued.
    Am i substituting or differentiating?
     
  7. Sep 19, 2007 #6

    arildno

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    Eeh?

    You are to show that by SUBSTITUTING into the diff.eq, which requires you to perform differentiation twice, the differential equation reduces to to an IDENTITY, valid for all t, i.e, that the given function is, indeed, a SOLUTION to the differential equation.
     
  8. Sep 19, 2007 #7
    So I am differentiating

    [tex]\ddot{\phi}[/tex]= -w[tex]^{2}[/tex](Asin(wt) + Bcos(wt))
     
  9. Sep 19, 2007 #8

    arildno

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    Quite so!

    Does that equal [itex]-w^{2}\phi[/itex]??
     
  10. Sep 19, 2007 #9
    I hope so, it probably will turn out not to be so when I do it wrong :)
     
  11. Sep 19, 2007 #10
    (long derivative..)
     
  12. Sep 19, 2007 #11
    wt wouldn't be constants... not treated as normal variables?
     
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