Solving O.D.E. Given Sequence (c, s, 1, e_1, e_{-1})

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 4K views
VinnyCee
Messages
486
Reaction score
0
Problem:

The sequence (c, s, 1, [tex]e_1,\,e_{-1}[/tex]) is a basis for the solution space of some differential equation p(D)y = 0. Find this O.D.E.

NOTE: c = cos(t) and s = sin(t)


Work so far:

I know that [tex]e_1[/tex] gives a (t - 1) and that the [tex]e_{-1}[/tex] gives a (t + 1), but how do I solve for the 1! I think that the c and s give ([tex]t^2[/tex] - 1).

Also, can someone explain in detail or give a reference to what a Ker() is?

thanks
 
Physics news on Phys.org
Ker(x) means the kernel of the transformation x--the set of all values that x maps to the identity. In linear algebra that would be the set of all values that x maps to the zero vector. What exactly are [tex]e_1,\,e_{-1}[/tex]?
 
VinnyCee said:
Problem:

The sequence (c, s, 1, [tex]e_1,\,e_{-1}[/tex]) is a basis for the solution space of some differential equation p(D)y = 0. Find this O.D.E.

NOTE: c = cos(t) and s = sin(t)


Work so far:

I know that [tex]e_1[/tex] gives a (t - 1) and that the [tex]e_{-1}[/tex] gives a (t + 1), but how do I solve for the 1! I think that the c and s give ([tex]t^2[/tex] - 1).
I have no idea what you mean by this! Is it possible that [tex]e_1[/tex] and [tex]e_{-1}[/tex] were supposed to be [itex]e^1[/itex] and [tex]e^{-1}[/itex]? If that is the case then the roots of the characteristic equation are i, -i, 0, 1, and -1. From that information, you should be able to find the characteristic equation and from that the differential equation.<br /> <br /> The problem as given ("Find this O.D.E.") has no single solution. There exist an infinite number of differential equations having those functions as solutions. Why I am giving is the simplest linear, homogenous, differential equation.[/tex]
 
I think he meant [tex]e^t[/tex] and [tex]e^{-t}[/tex] since c, s, and 1 are functions (well, e and 1/e would be functions too, but for these purposes would be equivalent to 1)