Mathematics 263                                  Laboratory 1                               September 1, 1993
	  
Name: _____________________________   Lab Partner: ___________________________
	  
Consulted with:  ____________________________________________________________

Introduction to Second Order
Differential Equations

In this laboratory we will learn how to solve second order differential equations that are linear with constant coefficients, that is equations of the form

where a, b, and c are constants and a is not 0 (if the equation is really second order).

Part 1:  Homogeneous Equations

The word homogeneous is overworked in the sciences and mathematics, but in this course a "homogeneous equation" is an equation that zero will solve.  In the space below (after lab period is over?), show that the differential equation above is homogeneous if and only if g(t) = 0.

Let's attempt to solve the homogeneous differential equation

The differential operator in this differential equation is

L[ y_ ] := D[y, {t, 2}] + 5 D[y, t] + 4 y

Define this equation in Mathematica by entering the command

diffeq = L[ y[t] ] == 0

We begin by noticing that the equation will be solved by a function if a sum of multiples of the function and its 1st and 2nd derivatives is zero.  So, it makes sense to consider functions whose derivatives are similar to the original function.  Accordingly, define

f[t_] := E^(r t)

Write this function and its 1st and 2nd derivatives in standard mathematical notation below, and explain why it is appropriate to use it here.

We can substitute this "template" function into the equation by using the command:

feq = diffeq /. y -> f

In your own words, briefly explain what this command does:

Notice that there is a common factor of ert on the left side of this equation.  We can divide it out as follows:

feq2 = feq[[1]]/E^(r t) == feq[[2]]/E^(r t)

This is called the characteristic (in Rob's terminology) or auxiliary (in Swokowski's) equation.  If we simplify it we find that it is just a polynomial equation in the variable r!

chareq = Simplify[feq2]

Of course solving this equation is easy, but we can ask Mathematica to do it, anyway.

charsoln = Solve[chareq,r]

This tells us that there are two values of r that yield solutions of the original differential equation: r = –4 and r = –1.  We can define the two solutions

y1[t] := f[t] /. charsoln[[1]];
y2[t] := f[t] /. charsoln[[2]];

We can check that the definitions are correct

{y1[t], y2[t]}

and check that they solve the original differential equation.

{L[ y1[t] ], L[ y2[t] ]}

Now we have two solutions to the differential equation, but what is the general solution (the form for all solutions)?

Part 2:  The Superposition Principle

Check to see which of the following functions solve the differential equation.  You may do this by hand or use Mathematica, as the command below suggests.  Indicate which ones solve and which don't.

a)  y1(t) + y2(t)                                  d) –3 y1(t) + 2y2(t)

b)  y1(t) – y2(t)                                  e)  y1(t)/ y2(t)

c)  y1(t) y2(t)                                      f) 4 y1(t) + 0.5y2(t)

L[ y1[t] + y2[t] ] //Simplify

Make a hypothesis about the form of the general solution of the differential equation.

The general solution of the equation has the form:

Verify your general solution satisfies the differential equation (by hand or with Mathematica).

If your hypothesis is correct, it is is an example of the Superposition Principle, which says:  a linear differential operator L has the property that L[C1y1 + C2y2] = C1L[y1] + C2L[y2].

Part 3:  Show What You Know

Use the technique you have learned to find a general solution for the differential equation

Plot the particular solution ( y1(t) + y2(t))/2 using the command below.  What sort of function is this?  Briefly explain how this function appeared here (Friday's lecture should help!).

Plot[(y1[t] + y2[t])/2, {t,0,2}]