Variation of parameters to obtain PS of 2nd Order non-hom equation

  • Thread starter robot1000
  • Start date
  • #1
5
0
The question I'm trying to solve is:

y" - 6y' + 9y = [itex]\frac{exp(3x)}{(1+x)}[/itex]

I formulated the Gen solution which are:

y1(x) = exp(3x) and y2(x) = xexp(3x)

I've then calculated the wronskian to get: exp(6x)

I then went onto to use the variation of parameters formula, which is where I got a bit stuck

eq0027M.gif


I ended up with

-exp(3x)*(x - ln(x+1) + xexp(3x)*ln(1+x)

The problem is, it just doesn't look right.

I would appreciate some guidance with this problem
 

Answers and Replies

  • #2
S_Happens
Gold Member
305
3
Except for a parentheses that you missed to close the (x - ln(x+1)), it looks right to me. Why would you say that it doesn't look right, and what guidance do you expect to get?
 
  • #3
1,796
53
I would appreciate some guidance with this problem
Here's what you do. You solve it numerically first and then plot the analytic solution you get over the numeric solution. If they agree, right on top of one another, then there is very good odds your analytic solution is correct. If you're going to work with DEs, this is a very useful practice in my opinion. So learn how to set all this up in Mathematica or another CAS and you will never say again, "that don't look right."
 

Related Threads on Variation of parameters to obtain PS of 2nd Order non-hom equation

Replies
1
Views
671
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
6
Views
995
Replies
2
Views
2K
  • Last Post
Replies
6
Views
4K
Replies
0
Views
2K
Replies
1
Views
798
Replies
20
Views
6K
Replies
11
Views
2K
Top