Variation of Parameters

  • Thread starter cheeee
  • Start date
  • #1
15
0
Solve by method of variation of parameters
(x^2)y'' - (4x)y' + 6y = x^4*sinx (x > 0)

Hey, I know how to solve problems using variation of parameters but only when the corresponding homogenous equation has constant coefficients...

y'' - (4/x)y' + (6/x^2)y = 0.. the bit im confused about is how to obtain the fundamental solutions to this equation {y1, y2} when the coefficients are not constants. Any help would be appreciated.

Thanks.
 

Answers and Replies

  • #2
StatusX
Homework Helper
2,564
1
Note that each term has the same "units", if you think of x as having units of length (so that a derivative removes one unit of length). Such differential equations have solutions of the form x^r. Plug this in and solve for r, and you'll quickly see why such solutions work. This is something you can just remember, although it would also fall out if you tried the Frobenius method.
 
  • #3
15
0
im not quite sure I understand where to plug in x^r .
 
  • #4
StatusX
Homework Helper
2,564
1
I mean y(x)=xr is a solution to the homogenous differential equation for certain r. Plug in this y and see which r work.
 

Related Threads on Variation of Parameters

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
868
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
832
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
13
Views
1K
Top