Variation of Parameters

  • Thread starter newtomath
  • Start date
  • #1
37
0
Given t^2 y'' -t(t+2)y' = (t+2)y= 2t^3 and y1= t, y2= te^t
Find the particular solution-

I ve worked the problem to [ -2t^2 -2t] by:
-t * Integral [ 2t* te^t/ t^2e^t] + te^t * Integral [ 2t^2/ t^2e^t]


whereas the book states that it is simply -2t^2. Can you guys tell me where I made my mistake, I am stumped.
 

Answers and Replies

  • #2
479
32
Given t^2 y'' -t(t+2)y' = (t+2)y= 2t^3 and y1= t, y2= te^t
Is that first = supposed to be +?
 
  • #3
459
0
Given t^2 y'' -t(t+2)y' = (t+2)y= 2t^3 and y1= t, y2= te^t
Find the particular solution-

I ve worked the problem to [ -2t^2 -2t] by:
-t * Integral [ 2t* te^t/ t^2e^t] + te^t * Integral [ 2t^2/ t^2e^t]

whereas the book states that it is simply -2t^2. Can you guys tell me where I made my mistake, I am stumped.
To be honest, I don't completely understand this, but I'm not sure you've actually done anything wrong. You got the particular solution -2t^2-2t, and the book says -2t^2. But notice that one of the homogeneous solutions is t. If you add a linear combination of the homogeneous solutions to the particular solution, you get another equally valid particular solution. So, if -2t^2 -2t works, so does -2t^2 -2t + 2t = -2t^2.
 
  • #4
MathematicalPhysicist
Gold Member
4,308
216
Well a private solution here will be of the form:
At^3+Bt^2+Ct
where you can see quite immediately that A=0 (cause it has a term of t^4 which obviously gets nullified).

I did the calculation and got exactly as the book, try again.
PS
the C term doesn't get cancelled though.
 
Last edited:

Related Threads on Variation of Parameters

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
687
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
2K
Replies
5
Views
4K
Replies
6
Views
1K
Top