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Discovery guide 27.1 Discovery guide

Discovery 27.1.
(a)

Verify that \(y(t) = 5 e^{-2t}\) solves the differential equation \(\dydt = -2 y(t)\text{.}\)

Note: Remember that verifying a function is a solution to a differential equation does not require you solve the differential equation — you just need to verify that left-hand and right-hand sides evaluate to the same result when the proposed solution function is substituted.

What is the initial value of the solution function?

(b)

Remind yourself of the pattern: the general solution to \(\dydt = k y\) is \(y(t) = \underline{\hspace{13.6363636363636em}}\text{.}\)

Discovery 27.2.

Consider the following coupled system of linear differential equations:

\begin{equation*} \left\{\begin{array}{rcrcr} \dxdt \amp = \amp 5 x(t) \amp - \amp 6 y(t) \text{,} \\ \dydt \amp = \amp 3 x(t) \amp - \amp 4 y(t) \text{.} \end{array}\right. \end{equation*}
(a)

Suppose we create new functions

\begin{equation*} \left\{\begin{array}{rcrcr} w(t) \amp = \amp - x(t) \amp + \amp y(t) \text{,} \\ z(t) \amp = \amp x(t) \amp - \amp 2 y(t) \text{,} \end{array}\right. \end{equation*}

out of the old. Create a new system of differential equations in \(w(t), w'(t), z(t), z'(t)\text{.}\)

What happened?

Hint

Differentiate the expression for \(w(t)\) to get a formula for \(w'(t)\) in terms of \(x'(t)\) and \(y'(t)\text{.}\) Then substitute the expressions for \(x'(t)\) and \(y'(t)\) from the original differential equations into your expression for \(w'(t)\text{.}\) Simplify, and then see if you can relate what you have back to the change-of-variable expressions for \(w(t)\) and \(z(t)\text{.}\) Then repeat for \(z(t)\text{.}\)

(b)

Now use the pattern from Task 27.1.b to solve the simplified system in the new variables \(w(t), z(t)\text{.}\)

(c)

If we want to convert these solutions for \(w(t), z(t)\) to solutions to for \(x(t),y(t)\text{,}\) we'll need to reverse the change of variables. That is, we'll need to solve the system

\begin{equation*} \left\{\begin{array}{rcrcr} - x \amp + \amp y \amp = \amp w \text{,} \\ x \amp - \amp 2 y \amp = \amp z \text{,} \end{array}\right. \end{equation*}

for \(x\) and \(y\text{.}\)

Wait! The pattern of the equations above looks familiar … Perhaps we could use linear algebra to solve them. (And maybe use matrix inversion to solve, instead of row reducing.)

Once you've solved the reverse change of variables, express the solutions for \(x(t)\) and \(y(t)\) as combinations of the solutions for \(w(t)\) and \(z(t)\text{.}\)

(d)

Task c demonstrated that the linear change-of-variable equations could be written in matrix form:

\begin{equation*} \begin{bmatrix} w \\ z \end{bmatrix} = \begin{bmatrix} \underline{\hspace{0.909090909090909em}} \amp \underline{\hspace{0.909090909090909em}} \\ \underline{\hspace{0.909090909090909em}} \amp \underline{\hspace{0.909090909090909em}} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\text{.} \end{equation*}

But the original system of differential equations involving \(x(t),y(t)\) also looks linear. Can you write that differential system in terms of matrix multiplication as well? One side of your matrix equation should involve a \(2 \times 2\) matrix times \(\left[\begin{smallmatrix} x \\ y \end{smallmatrix}\right]\text{.}\) Can you convert the other “differential” side of the matrix equation into an expression involving \(\left[\begin{smallmatrix} x \\ y \end{smallmatrix}\right]\text{?}\)

Can you also turn your simplified system involving \(w(t),w'(t),z'(t),z(t)\) into a matrix equation? What do you notice about coefficient matrix in this system?

(e)

So we have

  • a coefficient matrix relating \(\ddt \left[\begin{smallmatrix} x \\ y \end{smallmatrix}\right] \) to \(\left[\begin{smallmatrix} x \\ y \end{smallmatrix}\right] \text{,}\)
  • a coefficient matrix relating \(\ddt \left[\begin{smallmatrix} w \\ z \end{smallmatrix}\right] \) to \(\left[\begin{smallmatrix} w \\ z \end{smallmatrix}\right] \text{,}\) and
  • a coefficient matrix relating coordinate systems \(\left[\begin{smallmatrix} w \\ z \end{smallmatrix}\right] \) and \(\left[\begin{smallmatrix} x \\ y \end{smallmatrix}\right] \text{.}\)

What recent topic that we've been studying do you think relates these three coefficient matrices together?

Discovery 27.3.

Work out the pattern of Discovery 27.2. Suppose matrices \(A,B\) are similar via transition matrix \(P\) in the similarity relation \(\inv{P}AP = B\text{,}\) and that \(y_1(t), y_2(t), \dotsc, y_n(t)\) are functions that satisfy the differential matrix equation

\begin{align*} \ddt \uvec{y}(t) \amp = A \uvec{y}(t) \text{,} \amp \text{where } \uvec{y}(t) \amp = \begin{bmatrix} y_1(t) \\ y_2(t) \\ \vdots \\ y_n(t) \end{bmatrix} \text{.} \end{align*}

Substitute the similarity relation into the differential matrix equation and rearrange to get a new differential equation

\begin{equation*} \ddt \uvec{w}(t) = B \uvec{w}(t) \text{,} \end{equation*}

where \(\uvec{w}(t)\) is some change of variables from \(\uvec{y}(t)\text{.}\) Be explicit about how your change of variables relates \(\uvec{y}(t)\) and \(\uvec{w}(t)\text{.}\)

Discovery 27.4.

Discovery 27.2 demonstrated that if a differential matrix equation \(\ddt \uvec{y}(t) = A \uvec{y}(t) \) has a diagonalizable coefficient matrix \(A\text{,}\) then a change of variables via a transition matrix \(P\) that diagonalizes \(A\) will decouple the underlying system of equations, leaving simple proportional differential equations that are solved by exponential functions \(w_j(t) = c_j e^{k_j t}\text{.}\)

(a)

What do the constants \(k_j\) represent relative to the diagonal coefficient matrix \(\inv{P} A P\text{?}\) What do they represent relative to the original coefficient matrix \(A\text{?}\)

(b)

Suppose you were given a collection of initial values

\begin{equation*} \left\{\begin{array}{rcr} y_1(0) \amp = \amp a_1 \text{,} \\ y_2(0) \amp = \amp a_2 \text{,} \\ \amp \vdots \\ y_n(0) \amp = \amp a_n \text{.} \end{array}\right. \end{equation*}

We can collect these into an initial vector

\begin{equation*} \uvec{y}(0) = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} \text{.} \end{equation*}

How does this vector relate to the initial values for the \(w_j(t)\) solution functions?

Discovery 27.5.

The equation

\begin{equation*} y''(t) + 5 y'(t) + 4 y(t) = 0 \end{equation*}

is an example of a homogeneous, linear, second-order differential equation.

We will try to use vector/matrix methods to solve this equation. The vector of unknown functions we will use is

\begin{equation*} \mathbf{y}(t) = \begin{bmatrix} y(t) \\ y'(t) \end{bmatrix} \text{.} \end{equation*}

So, effectively we are setting \(y_1(t) = y(t)\) and \(y_2(t) = y'(t)\text{.}\)

(a)

Let's set up our system of linear differential equations.

\begin{align*} y_1'(t) \amp = \underline{\hspace{1.363636363636364em}} y_1(t) + \underline{\hspace{1.363636363636364em}} y_2(t) \\ y_2'(t) \amp = \underline{\hspace{1.363636363636364em}} y_1(t) + \underline{\hspace{1.363636363636364em}} y_2(t) \end{align*}
Hint

You can obtain the first equation by combining our definitions of \(y_1(t)\) and \(y_2(t)\text{.}\) For the second equation, use the original differential equation and the fact that

\begin{equation*} y_2'(t) = y_1''(t) \text{.} \end{equation*}
(b)

Solve the system from Task a by diagonalizing the coefficient matrix to decouple the system.

(c)

As part of Task b, you should have computed the characteristic polynomial of the coefficient matrix.

Compare this coefficient polynomial with the original second-order differential equation. Do you notice anything special?