Randolph College
Department of Mathematics and Computer Science

 

Section 3.5 - Solving Nonhomogeneous Differential Equations

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Today's lecture involves the solution to nonhomogeneous, linear, second-order differential equations with constant coefficients. That is, equations of the form $$ay'' + by' + cy = g(t).$$ Here, \(g\) is a function of the independent variable \(t\). We do this in two steps:
 
1. Solve the associated homogeneous differential equations $$ay'' + by' + cy = 0,$$ the same equation except with a zero on the right-hand side.
 
2. Find a particular solution to the non-homogeneous differential equation.
 
The first video covers this in a little more depth.

 

We have already spent a good deal of time discussing the solution to second-order, linear, homogeneous differential equations with constant coefficients. The form of such a solution depends on whether the characteristic polynomial has two distinct real solutions, a repeated real solution, or two conjugate complex solutions.

In order to find a particular solution to the nonhomogeneous equation, we use a method called the method of undetermined coefficients. In order to use such a method, we need the following chart:
$$\begin{array}{|l|l|}\hline g(t) & y_p(t) \\ \hline a_n t^n + a_{n-1}t^{n-1} + \ldots + a_1 t + a_0 & t^s(c_n t^n + c_{n-1}t^{n-1} + \ldots + c_1 t + c_0)\\ (a_n t^n + a_{n-1}t^{n-1} + \ldots + a_1 t + a_0)e^{\alpha t} & t^s(c_n t^n + c_{n-1}t^{n-1} + \ldots + c_1 t + c_0) e^{\alpha t}\\ (a_n t^n + a_{n-1}t^{n-1} + \ldots + a_1 t + a_0)\cos(\beta t) & t^s[(c_n t^n + c_{n-1}t^{n-1} + \ldots + c_1 t + c_0) \cos(\beta t) + (d_n t^n + d_{n-1}t^{n-1} + \ldots + d_1 t + d_0)\sin(\beta t)]\\ (a_n t^n + a_{n-1}t^{n-1} + \ldots + a_1 t + a_0)\sin(\beta t) & t^s[(c_n t^n + c_{n-1}t^{n-1} + \ldots + c_1 t + c_0) \cos(\beta t) + (d_n t^n + d_{n-1}t^{n-1} + \ldots + d_1 t + d_0)\sin(\beta t)]\\ \hline \end{array}$$ This chart can be extended. But this will be sufficient for our purposes.
 
So, for instance, if the right side of our nonhomogeneous differential equation had the form \(g(t) = t^2 - 10t + 4\), in other words, a polynomial of degree 2, then according to the first line of the table, the particular solution would have the following form: $$y_p(t) = t^s(c_2t^2 + c_1t + c_0),$$ a polynomial whose degree is also 2 multiplied by some power of \(t\). It's our job to compute the coefficients \(c_2\), \(c_1\), and \(c_0\). We choose \(s\) to be the smallest non-negative power so that none of the terms in $y_p$ are solutions to the associated homogeneous differential equation. The second video discusses the solution to the following differential equation: $$y'' - 3y' + 2y = t + 2.$$ Note that this is a second-order, linear, differential equation with constant coefficients, and that it is nonhomogeneous since the left side of the equation isn't zero. In this case \(g(t) = t+2\) and so our particular solution has the form $$y_p(t) = t^s(c_1t + c_0).$$

 

 

In the last video, we solve the equation $$y'' - 3y' + 2y = (t + 2)e^t.$$ In this example, we will use the second line of the above table, and will find that \(s = 0\) does not work but that \(s = 1\) does.
 

I have completed the following:

a.  Watched/taken notes on the first video.

b.  Watched/taken notes on the first two videos.

c.  Watched/taken notes on all three videos.

This is Princess Leia (left) and Princess Buttercup (right). They say hi.