Second order linear differential equations nonhomogeneous equations

[clairez93]

I could not get LaTex to format properly, so I typed out the question and my work using Microsoft Word's equation editor. Please see the attached document, apologies for any inconvenience! These problems involve the techniques for the method of undetermined coefficients and variation of parameters.

 

[Mark44]

Your particular solution y_p = Ae^-tcos(2t) + Be^-tsin(2t) turns out to be the general solution of the related homogeneous problem, y'' + 2y' + 5y = 0.

If you substitute y_p into the latter equation you should find that y_p'' + 2y_p' + 5y_p is identically equal to zero.

For the nonhomogeneous problem, the right side is 4e^-tcos(2t), so instead of choosing y_p = Ae^-tcos(2t) + Be^-tsin(2t) as you did, you should use y_p = Ate^-tcos(2t) + Bte^-tsin(2t) as your particular solution. Run this function through your differential equation and determine the parameters A and B.

The general solution to the nonhomogeneous problem will be
y_g = c₁e^-tcos(2t) + c₂e^-tsin(2t) + Ate^-tcos(2t) + Bte^-tsin(2t), using the values of A and B that you already found. Use the initial conditions to find c₁ and c₂.

The reason for doing things this way has to do with repeated roots of the characteristic equation. The explanation is a little lengthy, but I'll try to cover the high points as briefly as I can.

Your equation can be represented as (D² + 2D + 5)y = 4e^-tcos(2t). Here I am using operator notation, D, to represent the derivative with respect to t.

Notice the similarity between the operator notation, D² + 2D + 5, and the left side of the characteristic equation r² + 2r + 5. That's not an accident. The roots of the characteristic equation are r = -1 +/- 2i. Each of these values causes the expression r² + 2r + 5 to have a value of zero.

The associated function is e^{(-1 +/- 2i)t} = e^-te^+/-2it. Rather than working with this, it's more convenient to use e^-tcos(2t) and e^-tsin(2t).

When the operator D² + 2D + 5 is applies to either of these functions (in fact, any of the four), the result is 0. So this operator can be said to "annihilate" e^-tcos(2t) and e^-tsin(2t), as well as any linear combination of them.

That's why there is no linear combination of these two functions will serve as a particular solution of your nonhomogeneous problem.

Your nonhomog. problem can be written this way: (D² + 2D + 5)y = 4e^-tcos(2t). If we apply the same operator again, we get
(D² + 2D + 5)(D² + 2D + 5)y = (D² + 2D + 5)4e^-tcos(2t) = 0.

This turns the original nonhomogeneous, 2nd order diff. eqn into a 4th order, homogeneous problem. This time the characteristic equation is (r² + 2r + 5)² = 0, and now we have the same roots, only repeated.

Since there are four roots, and the equation is fourth order, we have to have four linearly independent solutions, not just the two we had before. On top of e^-tcos(2t) and e^-tsin(2t) that we had before, we can get two more independent solutions by multiplying each of these two by t.

That's pretty much it - hope this helps.