PDA

View Full Version : Differential Equation - Nonhomogenous Hermite's Polynomial


cse63146
May27-09, 09:36 PM
1. The problem statement, all variables and given/known data

Find the general solution up to degree 6 of y'' + ty' + y = e-2t

2. Relevant equations



3. The attempt at a solution

I know how to solve it for y'' + ty' + y = 0, but what do I do about the e-2t?

Mark44
May27-09, 09:40 PM
Are you doing the homogeneous equation by a power series method? If so, combine your series for y'', ty', and y on the one side and equate them to the series for e-2t.

cse63146
May27-09, 09:49 PM
am I supposed to use taylor's polynomials for the e-2t after?

Mark44
May28-09, 01:38 AM
That's what I would do for starters. After all, you're only concerned with terms up to degree 6.

Start with the assumption that y = a0 + a1t + a2t2 and so on up to degree 8 or so (to get degree 6 term in your second derivative. Calculate y' and y'' and multiply y' by t, then add them together and group them by like powers of t. On the right side, you'll have the power series for e-2t, which looks like 1 - 2t + (2t)2/2! + a few more terms to get to degree 6. Then equate both sides and solve for your coefficients.

cse63146
May28-09, 01:50 PM
So would each group of t be equal to degree 6 of e-2t power series.

For example

(....)t = degree 6 of e-2t
(....)t2 = degree 6 of e-2t
(....)t3 = degree 6 of e-2t

and so forth

or would it look like this:

(....)t = degree 1 of e-2t
(....)t2 = degree 2 of e-2t
(....)t3 = degree 3 of e-2t

and so forth

Mark44
May28-09, 02:15 PM
This one.

(....)t = degree 1 of e-2t
(....)t2 = degree 2 of e-2t
(....)t3 = degree 3 of e-2t
and so forth

Don't forget the degree 0 term on each side.