- #1
Kaguro
- 221
- 57
- Homework Statement
- Find the particular integral of the following DE:
y'' + y' + 3y = 5cos(2x+3)
- Relevant Equations
- The operator method.
I use the operator method here:
(D^2 + D+3)y = 5cos(2x+3)
## y = \frac{1}{D^2+D+3} 5cos(2x+3) ##
## \Rightarrow y= \frac{5}{-(2)^2+D+3}cos(2x+3) ##
## \Rightarrow y= \frac{5}{-4+D+3}cos(2x+3) ##
## \Rightarrow y= \frac{5}{D-1}cos(2x+3) ##
At this, if I revert back to write:
(D-1)y = 5cos(2x+3)
and so
y' - y = 5cos(2x+3)
Is this new equation entirely equivalent to the original one? No, I think not.
But if I continue with the operator method:
## \Rightarrow y= \frac{5(D+1)}{D^2-1}cos(2x+3) ##
## \Rightarrow y= \frac{5(D+1)}{-4-1}cos(2x+3) ##
## \Rightarrow y= -(D+1)cos(2x+3) ##
## \Rightarrow y= 2sin(2x+3) - cos(2x+3) ##
Which is correct.
Then what seems to be the problem here? Why can't I stop using the D-operator method and solve using other methods?
(D^2 + D+3)y = 5cos(2x+3)
## y = \frac{1}{D^2+D+3} 5cos(2x+3) ##
## \Rightarrow y= \frac{5}{-(2)^2+D+3}cos(2x+3) ##
## \Rightarrow y= \frac{5}{-4+D+3}cos(2x+3) ##
## \Rightarrow y= \frac{5}{D-1}cos(2x+3) ##
At this, if I revert back to write:
(D-1)y = 5cos(2x+3)
and so
y' - y = 5cos(2x+3)
Is this new equation entirely equivalent to the original one? No, I think not.
But if I continue with the operator method:
## \Rightarrow y= \frac{5(D+1)}{D^2-1}cos(2x+3) ##
## \Rightarrow y= \frac{5(D+1)}{-4-1}cos(2x+3) ##
## \Rightarrow y= -(D+1)cos(2x+3) ##
## \Rightarrow y= 2sin(2x+3) - cos(2x+3) ##
Which is correct.
Then what seems to be the problem here? Why can't I stop using the D-operator method and solve using other methods?