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

(Book) Differential & Integral Calculus by Piskunov.

Thanks,but I've got no possibility to take this book,I'm in Magdeburg,Germany,and there's no library in my company

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

thanks

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)

Ok thanks,at last everything is OK,but how to find C1,C2 out of initial conditions?I've forgoten

for examople y(0)'=1,y''(0)=1

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.

Thanks in advance,I got the disired.

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)???

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 ?

I hope so,and what will happened with sin(t),should we take the derivative also,tha -> cost???Am I right?

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

Thanks a lot for ur help!