A second order equation and its solutions

Phase, Frequency, Amplitude, and all that..

A second order equation and its solutions

Combinations and variants

On the last few pages we saw that the pair of functions $y(t)=\sin(t), x(t)=\cos(t)$ satisfy the coupled differential equations

$\frac{dx}{dt} = -y \] \[ \frac{dy}{dt} = x$

We also noticed that each of these functions also satisfy the second order differential equation

$\frac{d^2y}{dt} = -y$

(Even though this is written for $ y $ , the function $ x(t) $ also satisfies this relationship). We also found that the most general form of the solution would be a function

$f(t)= A \sin(t) + B \cos(t)$

which also satisfies this same equation. This function is referred to as a superposition of the more elementary functions, sine and cosine (which could be considered as the building blocks of the most general form of the solution). When we talked about oscillating moods in a pair of lovers, we noticed that the constants $ A, B $ could be determined by specifying initial conditions. Two pieces of information were used to solve for these constants, because there are two unknown constants. (This is true in general for a system of two first order equations, or a single second order equation.)

The same situation holds for solutions of the second order equation, but here there is only one "unknown function" being sought. (It is called $ y(t) $ in the equation, but remember that it need not be the function that we called y(t) before: any function fitting this relationship is, by definition, a solution.)

For your consideration:

(1) By taking the second derivative of the function $f(t)= A \sin(t) + B \cos(t)$ , show that it satisfies the second order differential equation.
(2) What sort of information do you think we could use to determine the constants ? We no longer have two functions, only one, so what initial conditions might we use?

Initial values for the second order equation

Suppose we specify that at time $ t=0 $

, the following situation holds:

$y(0) = 1, ~~~y'(0) = 0$

One of these conditions is related to the derivative of the function at time $ t=0 $ , which is

$y'(t) = f'(t)= A \cos(t) - B \sin(t)$

The other condition is on the value of the function itself. Together, the two conditions tell us that:

$1 = y(0) = f(0) = A \sin(0) + B \cos(0) = B \] \[ 0 = y'(0) = f'(0)= A \cos(0) - B \sin(0) = A$

Notice that by using the initial conditions, we have found values of the two constants, $ A=0, B=1 $ , so that the solution that satisfies both the differential equation and the initial conditions is

$y= f(t) = \cos(t)$

For your consideration:

(1) Suppose the initial conditions had been . What would the values of the constants have been in that case?

Variation on a theme

Suppose we had started out with a slightly different system of equations,

$\frac{dx}{dt} = -a y \] \[ \frac{dy}{dt} = b x$

This system of equations can be shown to be equivalent to the single second order equation

$\frac{d^2y}{dt} = -(a b) y$

This leads us to investigate the more general case in which a second order equation of the form

$\frac{d^2y}{dt} = -w^2 y$

is encountered. It is not difficult to show that the functions

$\sin(wt),~~~\cos(wt)$

would then make up the building blocks of the solution, which, in its most general form is

$f(t) = A \sin(wt) + B \cos(wt)$

For your consideration:

(1) By taking the derivative of the second equation in the new system of equations, and eliminating using the first equation, show that the above second order equation is obtained.
(2) Again by taking derivatives (and remembering the Chain Rule), show that $\sin(wt),~~\cos(wt)$ are solutions of the new equation.
(3) Show that $f(t) = A \sin(wt) + B\cos(wt)$ is also a solution.
(4) Find the solution that also satisfies