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4.1 - 4.2 Higher-Order ODEs

Most of what we know about 2nd-order extend to higher-order ODEs.

nth-order linear

\[ P_0(t) \frac{d^n y}{dt^n} + P_1(t) \frac{d^{n-1} y}{dt^{n-1}} + \dots + P_{n-1}(t) \frac{dy}{dt} + P_n(t)y = G(t) \]

In the form with leading coefficient 1:

\[ \frac{d^n y}{dt^n} + p_1(t) \frac{d^{n-1} y}{dt^{n-1}} + p_2(t) \frac{d^{n-2} y}{dt^{n-2}} + \dots + p_n(t)y = g(t) \]

The equation has a unique solution on some interval \( I \) where all \( p_1(t), p_2(t), \dots, p_n(t), g(t) \) are continuous and the interval \( I \) contains the initial \( t \) (\( t_0 \)).

  • 2nd-order: 2 initial conditions, usually \( y(t_0), y'(t_0) \) up to \( y' \)
  • nth-order: \( n \) initial conditions
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\[ \begin{cases} y(t_0) = y_0 \\ y'(t_0) = y_0' \\ y''(t_0) = y_0'' \\ \vdots \\ y^{(n-1)}(t_0) = y_0^{(n-1)} \end{cases} \] \( n \) of these

For example, \( t(t-1) y^{(4)} + e^t y'' + 4t^2 y = 0 \)

Initial conditions: \( y(2)=1, y'(2)=0, y''(2)=3, y'''(2)=-1 \)

Rewriting in standard form:

\[ y^{(4)} + \left[ \frac{e^t}{t(t-1)} \right] y'' + \left[ \frac{4t^2}{t(t-1)} \right] y = 0 \]
Continuous for \( t \neq 0, t \neq 1 \)
A horizontal number line for variable t with marked points at 0, 1, and a cross at t=2. An interval I is shown starting from 1 to infinity.
\( I: (1, \infty) \)
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Fundamental Solutions of \(n^{th}\)-Order Equations

\(n^{th}\)-order \(\rightarrow\) \(n\) fundamental solutions

\[ y_1, y_2, y_3, \dots, y_n \]

General solution: \[ y = c_1 y_1 + c_2 y_2 + c_3 y_3 + \dots + c_n y_n \]

\(c_1, c_2, \dots, c_n\) come from the \(n\) initial conditions

Existence of Constants

Given some initial conditions, can we always find the constants?

Suppose for a \(4^{th}\)-order eq. we have \(y_1, y_2, y_3, y_4\)

initial conditions \(y_0, y_0', y_0'', y_0'''\)

General solution: \[ y = c_1 y_1 + c_2 y_2 + c_3 y_3 + c_4 y_4 \]

at \(t_0\),

\[ \begin{aligned} c_1 y_1 + c_2 y_2 + c_3 y_3 + c_4 y_4 &= y_0 \\ c_1 y_1' + c_2 y_2' + c_3 y_3' + c_4 y_4' &= y_0' \\ c_1 y_1'' + c_2 y_2'' + c_3 y_3'' + c_4 y_4'' &= y_0'' \\ c_1 y_1''' + c_2 y_2''' + c_3 y_3''' + c_4 y_4''' &= y_0''' \end{aligned} \]
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\[ \begin{bmatrix} y_1 & y_2 & y_3 & y_4 \\ y_1' & y_2' & y_3' & y_4' \\ y_1'' & y_2'' & y_3'' & y_4'' \\ y_1''' & y_2''' & y_3''' & y_4''' \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} y_0 \\ y_0' \\ y_0'' \\ y_0''' \end{bmatrix} \]

a unique solution for \(c_1, c_2, c_3, c_4\) if

\[ \begin{vmatrix} y_1 & y_2 & y_3 & y_4 \\ y_1' & y_2' & y_3' & y_4' \\ y_1'' & y_2'' & y_3'' & y_4'' \\ y_1''' & y_2''' & y_3''' & y_4''' \end{vmatrix} \neq 0 \]

Wronskian \(W[y_1, y_2, y_3, y_4]\)

\(W \neq 0 \rightarrow\) the solutions are linearly independent

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Wronskian Example and Linear Independence

For example, the differential equation \( y''' + 2y'' - y' - 2y = 0 \) has solutions \( e^t, e^{-t}, e^{-2t} \).

Wronskian Calculation:

\[ W = \begin{vmatrix} e^t & e^{-t} & e^{-2t} \\ e^t & -e^{-t} & -2e^{-2t} \\ e^t & e^{-t} & 4e^{-2t} \end{vmatrix} \]

cofactor expansion

Pick any row or column to expand, here, we pick row 1:

\[ = e^t \begin{vmatrix} -e^{-t} & -2e^{-2t} \\ e^{-t} & 4e^{-2t} \end{vmatrix} - e^{-t} \begin{vmatrix} e^t & -2e^{-2t} \\ e^t & 4e^{-2t} \end{vmatrix} + e^{-2t} \begin{vmatrix} e^t & -e^{-t} \\ e^t & e^{-t} \end{vmatrix} \]\[ = e^t (-4e^{-3t} + 2e^{-3t}) - e^{-t} (4e^{-t} + 2e^{-t}) + e^{-2t} (e^0 + e^0) \]\[ = -2e^{-2t} - 6e^{-2t} + 2e^{-2t} = -6e^{-2t} \neq 0 \text{ for any } t \]

If \( W \) is not identically zero (not zero all the time) on some interval \( I \), then a unique solution exists throughout \( I \) and the fundamental solutions are linearly independent.

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Constant-Coefficient Homogeneous Equations

\[ a_1 y^{(n)} + a_2 y^{(n-1)} + \dots + a_n y = 0 \]

Just like for 2nd-order, the solutions are of the form \( e^{rt} \).

Characteristic Equation

\[ a_1 r^n + a_2 r^{n-1} + \dots + a_n = 0 \]

\( n \) roots \( \rightarrow n \) \( e^{rt} \) solutions

Example

For example, \( y''' - 3y'' + 3y' - y = 0 \)

Char. eq: \( r^3 - 3r^2 + 3r - 1 = 0 \) (Cubic)

One root is a factor of the constant term:

-1 has factors 1, -1. One of these is a root.

Testing Roots

Is \( r = 1 \) a root?

\( 1 - 3 + 3 - 1 = 0 \) true, so \( r = 1 \) is a root.

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Solving Characteristic Equations

We begin with the cubic characteristic equation:

\[ r^3 - 3r^2 + 3r - 1 = (r - 1)(ar^2 + br + c) \]

To find the quadratic factor, we use polynomial long division.

Polynomial Long Division

\[ \begin{array}{rll} r^2 - 2r + 1 & \\ r - 1 \enspace \overline{| \enspace r^3 - 3r^2 + 3r - 1} & \\ - (r^3 - r^2) \phantom{000000} & \\ \hline -2r^2 + 3r \phantom{000} & \\ - (-2r^2 + 2r) \phantom{0} & \\ \hline r - 1 & \\ \end{array} \]

The result of the division is the quadratic factor:

\[ r^2 - 2r + 1 \]

Finding the Roots

The factored equation becomes:

\[ (r - 1)(r^2 - 2r + 1) = 0 \]
\[ (r - 1)(r - 1)(r - 1) = 0 \]

This gives us a triple root:

\[ r = 1, 1, 1 \]

General Solution

Applying the same repeated roots rule, the fundamental solutions are:

\[ y_1 = e^t, \quad y_2 = te^t, \quad y_3 = t^2e^t \]
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Higher-Order Differential Equation

Consider the 6th-order differential equation:

\[ y^{(6)} - y'' = 0 \]

The characteristic equation is:

\[ r^6 - r^2 = 0 \]

Factoring the expression:

\[ r^2(r^4 - 1) = 0 \]
\[ r^2(r^2 + 1)(r^2 - 1) = 0 \]

Roots and Fundamental Solutions

The roots of the characteristic equation are:

\[ r = 0, 0, 1, -1, i, -i \]

These roots correspond to the following fundamental solutions:

  • For repeated roots \( r = 0, 0 \): \( y_1 = 1, y_2 = t \)
  • For real roots \( r = 1, -1 \): \( y_3 = e^t, y_4 = e^{-t} \)
  • For complex roots \( r = i, -i \): \( y_5 = \cos(t), y_6 = \sin(t) \)

General Solution

The general solution is the linear combination of the fundamental solutions:

\[ y = c_1 + c_2t + c_3e^t + c_4e^{-t} + c_5\cos(t) + c_6\sin(t) \]