Solve second order differential equation

  • Thread starter Oblio
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  • #1
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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!
 

Answers and Replies

  • #2
arildno
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What does the word "substitution" normally mean?
 
  • #3
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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...
 
  • #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?
 
  • #5
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That's why I'm confsued.
Am i substituting or differentiating?
 
  • #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.
 
  • #7
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So I am differentiating

[tex]\ddot{\phi}[/tex]= -w[tex]^{2}[/tex](Asin(wt) + Bcos(wt))
 
  • #8
arildno
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Quite so!

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

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