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4.1 + 4.2 Higher-Order Linear Equations

Recall

The first-order linear equation:

\[ y' + p(t)y = g(t), \quad y(t_0) = y_0 \]

has a unique solution on an interval where \( p(t) \) and \( g(t) \) are continuous and containing \( t_0 \).

The second-order linear equation:

\[ y'' + p(t)y' + q(t)y = g(t), \quad y(t_0) = y_0, \quad y'(t_0) = y'_0 \]

has a unique solution on an interval where \( p(t) \), \( q(t) \), and \( g(t) \) are continuous and containing \( t = t_0 \).

For \( n^{\text{th}} \)-order linear eq.

\[ y^{(n)} + p_1(t)y^{(n-1)} + p_2(t)y^{(n-2)} + \dots + p_n(t)y = g(t) \]\[ y(t_0) = y_0, \quad y'(t_0) = y'_0, \quad y''(t_0) = y''_0, \dots, y^{(n-1)}(t_0) = y_0^{(n-1)} \]

has a unique solution on an interval where \( p_1, p_2, \dots, p_n \) and \( g(t) \) are continuous, containing \( t = t_0 \).

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Example

\[ y^{(5)} + \frac{t^2}{t-8} y''' + \frac{\sin t}{t-8} y'' + \frac{\ln t}{t-8} y = \frac{\sqrt{t}}{t-8} \]

Conditions for continuity:

  • \( t \neq 8 \)
  • \( t > 0 \)

The following number line illustrates the domain restrictions for the coefficients and the non-homogeneous term:

A horizontal number line with a mark at 0 and an X at 8, showing intervals of continuity.

Intervals where there are unique solutions:

\( (0, 8) \) or \( (8, \infty) \)

Need \( t_0 \) to pick.

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Solution of Constant-Coeff eq.

Example

\[2y''' - 4y'' - 2y' + 4y = 0\]

Characteristic eq:

\[2r^3 - 4r^2 - 2r + 4 = 0\]\[r^3 - 2r^2 - r + 2 = 0\]\[r^2(r - 2) - (r - 2) = 0\]\[(r - 2)(r^2 - 1) = 0\]

The roots are:

\[r = -1, 1, 2\]
\[y = c_1 e^{-t} + c_2 e^t + c_3 e^{2t}\]
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Example

\[y^{(4)} + 4y''' + 4y'' = 0\]\[r^4 + 4r^3 + 4r^2 = 0\]\[r^2(r^2 + 4r + 4) = 0\]\[r^2(r + 2)^2 = 0\]

The roots are:

\[r = 0, 0, -2, -2\]
\[y = c_1 + c_2 t + c_3 e^{-2t} + c_4 t e^{-2t}\]

Example

\[y''' - 5y'' + 3y' + y = 0\]\[r^3 - 5r^2 + 3r + 1 = 0\]

By inspection, \(r = 1\) is a root.

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

Factor by grouping is harder.

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\[ ar^3 + br^2 + cr - ar^2 - br - c = r^3 - 5r^2 + 3r + 1 \]

By comparing coefficients:

\( a = 1 \)
\( b - a = -5 \)
\( c - b = 3 \)
\( -c = 1 \)
\( b = -4 \)
\( c = -1 \)

The characteristic equation factors as:

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

Solving for the roots:

\( r = 1 \)
\[ r = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5} \]

General Solution

\[ y = C_1 e^t + C_2 e^{(2+\sqrt{5})t} + C_3 e^{(2-\sqrt{5})t} \]
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Example

An 8th-order DE has the following roots to the characteristic eq.

\[ r = 2, \, 3, \, 3, \, 3, \, 2+3i, \, 2-3i, \, 2+3i, \, 2-3i \]

What is the general solution?

\[ \begin{aligned} y = & C_1 e^{2t} + C_2 e^{3t} + C_3 t e^{3t} + C_4 t^2 e^{3t} \\ & + C_5 e^{2t} \cos 3t + C_6 e^{2t} \sin 3t \\ & + C_7 t e^{2t} \cos 3t + C_8 t e^{2t} \sin 3t \end{aligned} \]
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Example: Solving a Third-Order Differential Equation

Consider the following third-order linear homogeneous differential equation with initial conditions:

\[ y''' + y' = 0 \quad \text{with} \quad y(0)=0, \, y'(0)=1, \, y''(0)=2 \]

1. Characteristic Equation

First, we find the characteristic equation by substituting \( y = e^{rt} \):

\[ r^3 + r = 0 \]

Factoring the equation:

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

The roots are:

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

2. General Solution

The general solution is a linear combination of the solutions corresponding to each root:

\[ y = C_1 + C_2 \cos t + C_3 \sin t \]

3. Applying Initial Conditions

We use the initial conditions to solve for the constants \( C_1, C_2, \) and \( C_3 \).

For \( y(0) = 0 \):

\[ 0 = C_1 + C_2 \]

Differentiating to find \( y' \):

\[ y' = -C_2 \sin t + C_3 \cos t \]

For \( y'(0) = 1 \):

\[ 1 = C_3 \]

Differentiating again to find \( y'' \):

\[ y'' = -C_2 \cos t - C_3 \sin t \]

For \( y''(0) = 2 \):

\[ 2 = -C_2 \implies C_2 = -2 \]

Substituting \( C_2 = -2 \) back into the first equation:

\[ C_1 = 2 \]

Final Particular Solution

\[ y = 2 - 2 \cos t + \sin t \]