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Old   January 28, 2005, 05:25
Default Differential eq.
  #1
Viktor Bessonov
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Would u be so kind to help a stupid student.How to solve the deff.eq.:

y(t)''+A*y(t)'-B*sin(t)=0 Boundary codition you can envent yourself
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Old   January 28, 2005, 06:45
Default Re: Differential eq.
  #2
pkm
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(Book) Differential & Integral Calculus by Piskunov.
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Old   January 28, 2005, 07:03
Default Re: Differential eq.
  #3
Viktor Bessonov
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Thanks,but I've got no possibility to take this book,I'm in Magdeburg,Germany,and there's no library in my company
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Old   January 28, 2005, 07:57
Default Re: Differential eq.
  #4
ulysses
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y(t) = C*exp(-sqrt(-A)*t) + D*exp(sqrt(-A)*t)

+ B*sin(t)/(A-1)

of course if A<0 and A=/=1

C & D can be found using the initial conditions
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Old   January 28, 2005, 08:23
Default Re: Differential eq.
  #5
Viktor Bessonov
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thanks
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Old   January 28, 2005, 09:26
Default Re: Differential eq.
  #6
Tom
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The solution is, give or take a slip in the algebra,

t = C + D.exp(-At) + E.sin(t) + F.cos(t)

where C & D are constants fixed by the initial condition and

E= B/(1+A^2), F = AB/(1+A^2)
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Old   January 28, 2005, 09:47
Default Re: Differential eq.
  #7
Viktor Bessonov
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Ok thanks,at last everything is OK,but how to find C1,C2 out of initial conditions?I've forgoten
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Old   January 28, 2005, 09:51
Default Re: Differential eq.
  #8
Viktor Bessonov
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for examople y(0)'=1,y''(0)=1
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Old   January 28, 2005, 09:57
Default Re: Differential eq.
  #9
Tom
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You have, at t=0 values of y and y', write down the formulas for y and y' at t=0 and solve the equations. In your case D is determined trivially from the condition on y' substitute this into the condition for y to determine C.
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Old   January 28, 2005, 10:12
Default Re: Differential eq.
  #10
Viktor Bessonov
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Thanks in advance,I got the disired.
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Old   January 28, 2005, 10:20
Default Re: Differential eq.
  #11
Viktor Bessonov
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And how will be looking this eq. if I change it from the second order dif.eq. to a system of two eq. of first order?

dy(1)=y(2)

dy(2)=-A*y(2)+B*sin(t)???
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Old   January 28, 2005, 10:39
Default Re: Differential eq.
  #12
ulysses
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you're going to apply Runge Kutta methods , aren't you ?

so you set y=z' and if Y = [ y , z] is the vector solution, you have then the system

Y'= [ z , z' ]

or in other way

z = y' z'= y'' = -A*y + B sin(t)

Am I right ?
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Old   January 28, 2005, 11:01
Default Re: Differential eq.
  #13
Viktor Bessonov
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I hope so,and what will happened with sin(t),should we take the derivative also,tha -> cost???Am I right?
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Old   January 31, 2005, 04:59
Default Re: Differential eq.
  #14
ulysses
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well once you have Y' = F(Y,T) with F(Y,t) the vector function : F(Y,t)= [ y' ; -A*y + B*sin(t)]

if you discretize with Euler method (for the sake of simplicity) :

Y_n+1 = Y_n + Dt*F(Y_n,t) with initial values Y_0= [ y(t=0) ; z(t=0)=y'(t=0) ]

So you needn't to calculate any more derivatives.

hope I'm right
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Old   January 31, 2005, 05:21
Default Re: Differential eq.
  #15
Viktor Bessonov
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Thanks a lot for ur help!
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