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Problem 6

\[ \pi^{-1+2i} \]

Using the identities \( e^{\ln x} = x \) and \( e^{a+ib} = e^a(\cos b + i \sin b) \):

\[ = (e^{\ln \pi})^{-1+2i} \] \[ = e^{-\ln \pi} e^{i(2 \ln \pi)} = e^{\ln \pi^{-1}} (\cos(2 \ln \pi) + i \sin(2 \ln \pi)) \] \[ = \frac{1}{\pi} (\cos(2 \ln \pi) + i \sin(2 \ln \pi)) \]

Problem 23

\[ 3u'' - u' + 2u = 0, \quad u(0)=2, \, u'(0)=0 \]

Characteristic equation:

\[ 3r^2 - r + 2 = 0 \] \[ r = \frac{1 \pm \sqrt{1 - 4(3)(2)}}{2(3)} = \frac{1 \pm i\sqrt{23}}{6} \]

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3.4 Repeated Roots; Reduction of Order

\[ y'' + 6y' + 9y = 0 \] \[ r^2 + 6r + 9 = 0 \implies (r+3)(r+3) = 0 \] \[ r_1 = -3, \, r_2 = -3 \]

First solution: \( y_1 = e^{-3t} \)

2nd solution: \( y_2 = ? \) can't use \( y_2 = y_1 \) because Wronskian = 0 for all \( t \)

To find \( y_2 \), use method of Reduction of Order

Assume \( y_2 = v(t) y_1 \) where \( v(t) \) is a function of \( t \).

\[ y_2 = v e^{-3t} \]

Product rule:

\[ y_2' = -3v e^{-3t} + v' e^{-3t} \] \[ y_2'' = 9v e^{-3t} - 6v' e^{-3t} + v'' e^{-3t} \]

Plug \( y_2 \) into DE

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Solving Second-Order Linear Homogeneous Differential Equations

Sub into \( y'' + 6y' + 9y = 0 \)

\[ 9ve^{-3t} - 6v'e^{-3t} + v''e^{-3t} - 18v'e^{-3t} + 6v'e^{-3t} + 9ve^{-3t} = 0 \]

After simplification (canceling terms):

\[ v''e^{-3t} = 0 \]

Since \( e^{-3t} \neq 0 \), we have:

\[ v'' = 0 \]

\[ v' = c_1 \]

\[ v = c_1 t + c_2 \]

Finding the Second Solution

\[ y_2 = v y_1 = (c_1 t + c_2) e^{-3t} \]\[ y_2 = c_1 t e^{-3t} + c_2 e^{-3t} \]

For simplicity, choose \( c_1 = 1 \)

\( y_2 = c_1 t e^{-3t} \)

Repeat of \( y_1 \), don't include in \( y_2 \). Choose \( c_2 = 0 \)

\( c_2 e^{-3t} \)

Thus, the second solution is:

\[ y_2 = t y_1 \]

\[ y_2 = t e^{-3t} \]

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For 2nd order linear homogeneous, constant coefficient DE,

\( v \) is always \( t \)

\[ y'' + ay' + by = 0 \quad \leftarrow \text{any of this form w/ repeated roots.} \]

Example

\( 9y'' - 12y' + 4y = 0 \)

\( y(0) = 2, \, y'(0) = -1 \)

\[ 9r^2 - 12r + 4 = 0 \]\[ (3r - 2)(3r - 2) = 0 \]\[ r = 2/3, \, 2/3 \]

\( y_1 = e^{2/3 t} \)

\( y_2 = t e^{2/3 t} \)

Because DE is linear, homogeneous constant coeff

General Solution and Derivative

\[ y = c_1 e^{2/3 t} + c_2 t e^{2/3 t} \]\[ y' = \frac{2}{3} c_1 e^{2/3 t} + c_2 \left( \frac{2}{3} e^{2/3 t} \cdot t + e^{2/3 t} \right) \]

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\[ 2 = C_1 \]\[ -1 = \frac{2}{3}(2) + C_2 \quad \implies \quad C_2 = -\frac{7}{3} \]

\[ y(t) = 2e^{2/3t} - \frac{7}{3}te^{2/3t} \]

If \( t \) is small, first term (\( y_1 \)) dominates.

Solution looks like \( y_1 \) first.

If \( t \) is large, 2nd term (\( y_2 \)) is dominant.

Solution looks like \( y_2 \) later.

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Example

\[ t^2 y'' + 2ty' - 2y = 0 \]

linear, homogeneous

but not constant coeff.

\( y_1 = t \), find \( y_2 \) using Reduction of Order.

\( v(t) \neq t \) in general

\[ y_2 = v(t) y_1 = vt \]\[ y_2' = v + v't \]\[ y_2'' = v' + v' + v''t = 2v' + v''t \]\[ t^2(2v' + v''t) + 2t(v + v't) - 2vt = 0 \]\[ 2t^2v' + v''t^3 + 2tv + 2v't^2 - 2vt = 0 \]

Note: The terms \( 2tv \) and \( -2vt \) cancel out.

\[ t^3 v'' + 4t^2 v' = 0 \]\[ \rightarrow t^3 (v')' + 4t^2 (v') = 0 \]

\[ (v')' = -\frac{4t^2}{t^3} (v') \]

1st order in \( v' \)

linear

separable

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Solving for the Second Solution via Reduction of Order

\[\frac{d(v')}{dt} = -\frac{4}{t}(v')\]

Separating variables to solve for \(v'\):

\[\frac{1}{(v')} d(v') = -\frac{4}{t} dt\]

Integrating both sides:

\[\ln |v'| = -4 \ln t + k_1\]\[\ln |v'| = \ln t^{-4} + k_1\]

Exponentiating to find \(v'\):

\[v' = k_1 t^{-4}\]

Integrating \(v'\) to find \(v\):

\[v = \frac{k_1}{-3} t^{-3} + k_2 = k_1 t^{-3} + k_2\]

Determining the Second Solution \(y_2\)

Given the first solution \(y_1 = t\), we define the second solution as \(y_2 = vt\):

\[y_2 = (k_1 t^{-3} + k_2) t\]\[y_2 = k_1 t^{-2} + k_2 t\]

Analysis of terms:

The term \(k_2 t\) is already in \(y_1\). We don't want redundancy, so choose \(k_2 = 0\).
The term \(k_1 t^{-2}\) is not in \(y_1\). Choose \(k_1 = 1\) for simplicity.

Final Result

\(y_2 = t^{-2}\)