# Second order linear differential equations nonhomogeneous equations

• clairez93
In summary, the conversation discusses a problem involving the techniques for the method of undetermined coefficients and variation of parameters. It is recommended to use a particular solution of Ate-tcos(2t) + Bte-tsin(2t) for the nonhomogeneous problem, and to use initial conditions to find the values of c1 and c2. This approach is necessary due to the repeated roots of the characteristic equation. The conversation also discusses the operator notation and its relation to the characteristic equation, as well as the need for four linearly independent solutions in the fourth order problem.
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.

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• diffeq.doc
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Your particular solution yp = 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 yp into the latter equation you should find that yp'' + 2yp' + 5yp is identically equal to zero.

For the nonhomogeneous problem, the right side is 4e-tcos(2t), so instead of choosing yp = Ae-tcos(2t) + Be-tsin(2t) as you did, you should use yp = 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
yg = 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.

## 1. What is a second order linear differential equation?

A second order linear differential equation is a mathematical equation that involves a second derivative of an unknown function and can be written in the form of y'' + p(x)y' + q(x)y = r(x), where y is the unknown function, p(x) and q(x) are coefficients, and r(x) is a non-zero function.

## 2. What does it mean for a second order linear differential equation to be nonhomogeneous?

A nonhomogeneous second order linear differential equation is one that contains a non-zero function on the right-hand side, r(x). This means that the equation is not equal to zero and has a solution that is dependent on the function r(x).

## 3. How do you solve a second order linear differential equation?

To solve a second order linear differential equation, you can use techniques such as variation of parameters, undetermined coefficients, or the method of annihilators. These methods involve finding a particular solution to the nonhomogeneous equation and then finding the general solution by adding the solution to the corresponding homogeneous equation.

## 4. What is the difference between a homogeneous and nonhomogeneous second order linear differential equation?

A homogeneous second order linear differential equation only contains a function of y and its derivatives, while a nonhomogeneous equation includes an additional non-zero function on the right-hand side. This means that the solution to a homogeneous equation will be in the form of y = Aemx + Benx, while the solution to a nonhomogeneous equation will include a particular solution that is dependent on the non-zero function r(x).

## 5. When are second order linear differential equations used in real-world applications?

Second order linear differential equations are used in a variety of real-world applications, such as modeling the motion of a spring or pendulum, analyzing electrical circuits, and studying population growth or decay. They are also commonly used in engineering, physics, and economics to describe various phenomena and make predictions based on mathematical models.

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