 #1
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 18
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}{D1}cos(2x+3) ##
At this, if I revert back to write:
(D1)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^21}cos(2x+3) ##
## \Rightarrow y= \frac{5(D+1)}{41}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 Doperator 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}{D1}cos(2x+3) ##
At this, if I revert back to write:
(D1)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^21}cos(2x+3) ##
## \Rightarrow y= \frac{5(D+1)}{41}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 Doperator method and solve using other methods?