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7.1 First-order system of Diff. Eqs.

\[ x'(t) = f(x, y, t) \]\[ y'(t) = g(x, y, t) \]

x, y : dependent variables

t : independent variable

for example,

\[ \begin{cases} x' = y \\ y' = -x \end{cases} \]

in general, these equations are coupled → depend on other variable(s)

here, to solve for x, we need to know y and vice versa.

one way to solve a system like this is to convert it into a 2nd-order eq.

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\[ \begin{cases} x' = y \\ y' = -x \end{cases} \]

differentiate: \( x'' = y' \)

then from \( y' = -x \), we get \( x'' = -x \)

or \( x'' + x = 0 \) this we can solve

\[ r^2 + 1 = 0 \quad \implies \quad r = \pm i \]

so,

\[ x(t) = C_1 \cos t + C_2 \sin t \]

then from \( x' = y \) we get

\[ y = x'(t) = -C_1 \sin t + C_2 \cos t \]

in general, n 1st-order → one nth-order

and vice versa. (nth-order → n 1st-order)

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Converting Higher-Order ODEs to Systems

Example: Second-Order Equation

\[ x'' + 3x' + 7x = t^2 \]

2nd-order → system of 2 1st-order

Define two variables to represent \( x \) and \( x' \):

all dependent variables below the highest order of derivative

Let:

\[ x_1 = x \]\[ x_2 = x' \]

1st eq. of the system:

\[ x_1' = x_2 \]

2nd eq. of the system:

rewrite the given diff. eq. in terms of \( x_2' \)

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\[ x'' + 3x' + 7x = t^2 \]\[ \downarrow \quad \downarrow \quad \downarrow \]\[ x_2' + 3x_2 + 7x_1 = t^2 \]

So,

\[ x_2' = -3x_2 - 7x_1 + t^2 \]

Example: Fourth-Order Equation

\[ x^{(4)} + 6x''' - 3x'' + x' + 10x = \cos 3t \]

4th-order → 4 1st-order

Define 4 variables to represent \( x, x', x'', x''' \)

(all derivs below the highest one)
\[ x_1 = x \]\[ x_2 = x' \]\[ x_3 = x'' \]\[ x_4 = x''' \]
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Converting Higher-Order Differential Equations to Systems

Write a differential eq. for each:

\[ x_1' = x' = x_2 \]
\[ x_1' = x_2 \]
\[ x_2' = x_3 \]
\[ x_3' = x_4 \]

Final Equation Substitution

last one: the original diff. eq. rewritten in terms of the new variables

\[ x^{(4)} + 6x''' - 3x'' + x' + 10x = \cos 3t \]
\[ x_4' + 6x_4 - 3x_3 + x_2 + 10x_1 = \cos 3t \]
\[ x_4' = -6x_4 + 3x_3 - x_2 - 10x_1 + \cos 3t \]
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Order Reduction Principles

\( n^{\text{th}} \)-order \( \rightarrow n \) 1st-order

system of two 2nd-order \( \rightarrow 4 \) 1st-order

Example

\[ x'' - 5x' - 4x + 6y = 0 \]\[ y'' + 6y' + 5x + 5y = 0 \]

define new variables to represent \( x, x', y, y' \)

\[ z_1 = x \]\[ z_2 = x' \]\[ z_3 = y \]\[ z_4 = y' \]
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Converting to a System of First-Order Differential Equations

Write differential equations for each:

\( z_1' = z_2 \)

\( z_2' = x'' \) → look at the 1st 2nd-order eq.

\[ x'' - 5x' - 4x + 6y = 0 \] \[ x'' = 5x' + 4x - 6y \] \[ z_2' = 5z_2 + 4z_1 - 6z_3 \]

\( z_2' = 5z_2 + 4z_1 - 6z_3 \)

\( z_3' = z_4 \)

\( z_4' = y'' \) → the other 2nd-order eq.

\[ y'' + 6y' + 5x + 5y = 0 \] \[ y'' = -6y' - 5x - 5y \]

\( z_4' = -6z_4 - 5z_1 - 5z_3 \)

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Matrix Representation

We can rewrite the system as a matrix eq.

\[ \begin{bmatrix} z_1' \\ z_2' \\ z_3' \\ z_4' \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 4 & 5 & -6 & 0 \\ 0 & 0 & 0 & 1 \\ -5 & 0 & -5 & -6 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ z_3 \\ z_4 \end{bmatrix} \]
\( \vec{z}' = A \vec{z} \)
\( \vec{z} = \begin{bmatrix} z_1 \\ z_2 \\ z_3 \\ z_4 \end{bmatrix} \)

Solution of the system must have something to do with the matrix A

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We can see the connection in very system systems (more details in the rest of ch. 7)

\[ x'' - x = 0 \longrightarrow x(t) = c_1 e^t + c_2 e^{-t} \]

\[ r^2 - 1 = 0 \]

characteristic eq.

\[ \longrightarrow r = 1, -1 \]

Turn into a system:

Let \( z_1 = x \) and \( z_2 = x' \)

\[ \begin{cases} z_1' = z_2 \\ z_2' = z_1 \end{cases} \implies \begin{bmatrix} z_1' \\ z_2' \end{bmatrix} = \underbrace{ \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} }_{A} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} \]

Eigenvalues of A:

\[ \det(A - \lambda I) = 0 \]

\[ \begin{vmatrix} -\lambda & 1 \\ 1 & -\lambda \end{vmatrix} = 0 \]

\[ \lambda^2 - 1 = 0 \]

characteristic eq.

\[ \lambda = 1, -1 \]

  • For \( \lambda = 1 \), we have \( e^t \)
  • For \( \lambda = -1 \), we have \( e^{-t} \)