# Differential eq.

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 January 28, 2005, 05:25 Differential eq. #1 Viktor Bessonov Guest   Posts: n/a 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

 January 28, 2005, 06:45 Re: Differential eq. #2 pkm Guest   Posts: n/a (Book) Differential & Integral Calculus by Piskunov.

 January 28, 2005, 07:03 Re: Differential eq. #3 Viktor Bessonov Guest   Posts: n/a Thanks,but I've got no possibility to take this book,I'm in Magdeburg,Germany,and there's no library in my company

 January 28, 2005, 07:57 Re: Differential eq. #4 ulysses Guest   Posts: n/a 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

 January 28, 2005, 08:23 Re: Differential eq. #5 Viktor Bessonov Guest   Posts: n/a thanks

 January 28, 2005, 09:26 Re: Differential eq. #6 Tom Guest   Posts: n/a 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)

 January 28, 2005, 09:47 Re: Differential eq. #7 Viktor Bessonov Guest   Posts: n/a Ok thanks,at last everything is OK,but how to find C1,C2 out of initial conditions?I've forgoten

 January 28, 2005, 09:51 Re: Differential eq. #8 Viktor Bessonov Guest   Posts: n/a for examople y(0)'=1,y''(0)=1

 January 28, 2005, 09:57 Re: Differential eq. #9 Tom Guest   Posts: n/a 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.

 January 28, 2005, 10:12 Re: Differential eq. #10 Viktor Bessonov Guest   Posts: n/a Thanks in advance,I got the disired.

 January 28, 2005, 10:20 Re: Differential eq. #11 Viktor Bessonov Guest   Posts: n/a 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)???

 January 28, 2005, 10:39 Re: Differential eq. #12 ulysses Guest   Posts: n/a 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 ?

 January 28, 2005, 11:01 Re: Differential eq. #13 Viktor Bessonov Guest   Posts: n/a I hope so,and what will happened with sin(t),should we take the derivative also,tha -> cost???Am I right?

 January 31, 2005, 04:59 Re: Differential eq. #14 ulysses Guest   Posts: n/a 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

 January 31, 2005, 05:21 Re: Differential eq. #15 Viktor Bessonov Guest   Posts: n/a Thanks a lot for ur help!

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