Mixed method of solving differential equations

[Kaguro]

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?

 

[Mark44]

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)

$y = \frac{1}{D^2+D+3} 5cos(2x+3)$
$\Rightarrow y= \frac{5}{-(2)^2+D+3}cos(2x+3)$

I don't understand the above. Why did you replace $D^2$ with $-(2)^2$, but not also replace D? Where did $-(2)^2$ come from?

$\Rightarrow y= \frac{5}{-4+D+3}cos(2x+3)$
$\Rightarrow y= \frac{5}{D-1}cos(2x+3)$

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.

Right, they are different.

But if I continue with the operator method:
$\Rightarrow y= \frac{5(D+1)}{D^2-1}cos(2x+3)$

I don't understand this one, either. How did you come up with $\frac{5(D+1)}{D^2-1}$?

$\Rightarrow y= \frac{5(D+1)}{-4-1}cos(2x+3)$
$\Rightarrow y= -(D+1)cos(2x+3)$
$\Rightarrow y= 2sin(2x+3) - cos(2x+3)$

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?

 

[Kaguro]

https://arxiv.org/pdf/1802.09343
This paper describes the proofs of the operator method.

Well, this is how we work with trigonometric functions. This is called Eigenvalue substitution. The eigenvalue of D^2 operator on trig function cos(ax+b) is -a^2. So replace with that.

 

[Kaguro]

I don't understand this one, either. How did you come up with $\frac{5 (D+1)}{D^2-1}$?

We had:
$\frac{5}{D-1}cos(2x+3)$
Just multiply and divide by (D+1).

 

[Mark44]

I'm familiar with the annihilator method, which uses operators to solve linear differential equations, and have written a couple of Insights articles on this method - https://www.physicsforums.com/insights/solving-homogeneous-linear-odes-using-annihilators/ and https://www.physicsforums.com/insights/solving-nonhomogeneous-linear-odes-using-annihilators/. Problems similar to yours are discussed in the second article.

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.

No, it's not equivalent to the original differential equation, but it gives you a particular solution to it. IOW, it won't give you the general solution.
But, as your work shows, when you continue, you get a particular solution.

 

[Mark44]

Well, this is how we work with trigonometric functions. This is called Eigenvalue substitution. The eigenvalue of D^2 operator on trig function cos(ax+b) is -a^2.

Another way to look at this, is that the operator $(D^2 + a^2)$ annihilates $\cos(ax + b)$. In other words, $(D^2 + a^2)[\cos(ax + b)] = D^2(\cos(ax + b)) + a^2(\cos(ax + b)) = 0$.

 

[Kaguro]

I'm familiar with the annihilator method, which uses operators to solve linear differential equations, and have written a couple of Insights articles on this method - https://www.physicsforums.com/insights/solving-homogeneous-linear-odes-using-annihilators/ and https://www.physicsforums.com/insights/solving-nonhomogeneous-linear-odes-using-annihilators/. Problems similar to yours are discussed in the second article.

These articles proved extremely helpful to me. Thank you for suggesting me these.