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Differential equation Task 7

  1. Dec 13, 2013 #1
    Task 7
    Show that y=(1/4)tsin2t satisfies equation

    d2y/dt2+4y=cos2t

    Find the general solution and deduce the solution which satisfies y(0)=0 and y'(0)=0. What happens as t increases?

    Solution

    In the end I stay with:
    y=Acos2t+Bsin2t+(1/4)tsin2t

    dy/dx=-2Asin2t+2Bcos2t+(1/2)tcos2t+(1/4)sin2t

    After applying boundary conditions I get:
    0=0

    What does it mean and what happens as 't' is increasing? I was asked to plot a graph.
     
  2. jcsd
  3. Dec 13, 2013 #2

    ehild

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    Homework Helper
    Gold Member

    Yes, it is the general solution of the differential equation. Now you have to choose A and B so that both y and y' is zero at t=0.

    0=0 does not mean anything. What are A and B if y(0)=0 and y'(0)=0?

    Substitute zero for t in y=Acos2t+Bsin2t+(1/4)tsin2t, what do you get? Do the same for y'.

    You have to plot the y(t) graph for t>0.


    ehild
     
  4. Dec 13, 2013 #3
    I get that
    A=0
     
  5. Dec 13, 2013 #4
    How can I find B?
     
  6. Dec 13, 2013 #5
    If B equals 0 as well then the graph will look like a Christmas tree.

    But why would B equal 0?
     
  7. Dec 13, 2013 #6
    But this is simple!
    I had some brain freeze...
    Thanks for help!
     
  8. Dec 13, 2013 #7

    ehild

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    Gold Member

    Simple, isn't it? y'(0)=0, so -2Asin2t+2Bcos2t+(1/2)tcos2t+(1/4)sin2t=0 and A=0, 0=2Bcos(0) -->B=0.
    Yes, the graph looks like a horizontal Christmas-tree:smile:

    ehild
     
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