I am trying to set up the appropriate particular solution for this differential equation but don't know where to start. It doesn't even seem like a ode since I only have y terms on the LHS and no derivatives of y.
Any help would be appreciated just to set this up. Thanks
(D-1)^3(D^2-4)y=xe^x +e^2x +e^-2x
Hello, doke!
Are you in self-study?
This notation should have been explained to you in class.
When using the "Method of Operators", they often use this form of notation.
means . . . means . . . means
Example
An equation like can be written: .
Then we can factor out the y (ha!): .
To solve for , we would "divide by "
A really silly thing to say because has no meaning . . . or does it?
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Suppose we have the equation: .
We can solve it with an integrating factor: .
Multiply through by
The left side is the derivative of a product: .
Integrate: .
Therefore: .
The same equation can be written: .
. . which gives us: .
We have now defined that 'rather silly' statement:
. . . . .
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Back to our example . . .
We had: .
By definition: .
Integrate by parts and get: .
Therefore: .
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Now, your problem is inhumanely long and difficult!
You have: .
You have:
And you are expected to "divide by " three times
. . then "divide by ", then "divide by ".
. . . I'll wait in the car.
We need a particular solution to this. Note that:
To get a homogeneous solution for this differential equation, note that all of the coefficients of the D operator are constants. This means that the homogeneous solutions are going to be of the form:
where the a, b, ... solve the characteristic polynomial.
As to the solution of the characteristic polynomial, we know them, since the problem gives us a factored version of the operator. The characteristic polynomial is:
.
This has roots at the three cube roots of 1 and at m = 2, -2.
This suggests the following form for the particular solution:
We need the first term to produce the on the RHS of the differential equation. We need the second and third terms the way they are in order to produce the . We need to multiply by the extra "x" in this case because both and are already in the homogeneous solution.
I hope I explained that well enough!
-Dan