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3.3 Complex Roots of the Char. Eq.

\[y'' + y = 0 \quad r^2 + 1 = 0 \quad r = i, -i \quad i^2 = -1 \quad i = \sqrt{-1}\]

\(y\) such that it is the negative of its 2nd derivative

\(\rightarrow y = \cos(t), y = \sin(t)\)

Solution: \(y = e^{rt}\)

\[y_1 = e^{it} \quad y_2 = e^{-it}\]

\(\cos(t), \sin(t)\) somehow?

how does this work?

\[e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \frac{t^4}{4!} + \dots + \frac{t^n}{n!} + \dots\]
\[e^{(it)} = 1 + it - \frac{t^2}{2!} - \frac{it^3}{3!} + \frac{t^4}{4!} + \frac{it^5}{5!} + \frac{-t^6}{6!} + \dots\]
\[= \left( 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \frac{t^6}{6!} + \dots \right) + i \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \dots \right)\]

\(e^{it} = \cos(t) + i \sin(t)\) Euler's Formula

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\[y'' + y = 0 \quad r = \pm i\]
\[y_1 = e^{it} = \cos(t) + i \sin(t)\]
\[y_2 = e^{-it} = e^{i(-t)} = \cos(-t) + i \sin(-t)\]
\[= \cos(t) - i \sin(t)\]

general solution: \(y = c_1 y_1 + c_2 y_2\) is real for real-valued \(y(t_0) = y_0, y'(t_0) = y'_0\)

expression contains \(i\)

usually, we don't want \(i\) in our real-valued solutions

notice the real part and imaginary part (\(\cos(t), \sin(t)\)) are themselves solutions to \(y'' + y = 0\)

so, it's equally valid to say \(y_1 = \cos(t), y_2 = \sin(t)\)

\(\rightarrow\) use the real and imaginary parts of either solution as the fundamental solutions

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Example: Second-Order Linear Homogeneous ODE

Consider the differential equation:

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

The characteristic equation is:

\[ r^2 + 2r + 2 = 0 \]

Solving for \( r \) using the quadratic formula:

\[ r = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i \]

Finding Fundamental Solutions

Form one solution using \( e^{rt} \), then extract the real and imaginary parts as fundamental solutions.

\[ e^{(-1+i)t} = e^{-t} e^{it} = e^{-t} (\cos(t) + i \sin(t)) \]\[ = e^{-t} \cos(t) + i [ e^{-t} \sin(t) ] \]

So, the fundamental solutions are:

\[ y_1 = e^{-t} \cos(t) \]\[ y_2 = e^{-t} \sin(t) \]

General Solution

\[ y = c_1 e^{-t} \cos(t) + c_2 e^{-t} \sin(t) \]
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Visualizing the Solution

The following graph illustrates the behavior of the solution, showing the oscillation bounded by exponential decay envelopes.

Graph of a decaying oscillation y(t) bounded by exponential envelopes e^-t and -e^-t on t and y axes.
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3.4 Repeated Roots, Reduction of Order

\[y'' - 2y' + y = 0\]\[r^2 - 2r + 1 = 0 \quad (r-1)(r-1) = 0 \quad r = 1, 1\]

\(y_1 = e^{r_1 t} = e^t\)

\(y_2 = e^{r_2 t} = e^t ?\)

\[W[y_1, y_2] = \begin{vmatrix} e^t & e^t \\ e^t & e^t \end{vmatrix} = 0\]

cannot find \(c_1, c_2\) to meet the initial condition to find a unique solution

\(y_2\) must NOT be the same as \(y_1\)

\(y_2 = ?\)

Reduction of Order (due to D'Alembert)

\(y_2 = v(t) y_1\)

to find \(v(t)\), sub \(y_2 = v(t) y_1\) into the diff. eq.

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\[y'' - 2y' + y = 0\]\[y_1 = e^t\]\[y_2 = v e^t\]\[y_2' = v e^t + v' e^t\]\[y_2'' = v e^t + v' e^t + v' e^t + v'' e^t\]\[= v e^t + 2v' e^t + v'' e^t\]

Substituting into the differential equation:

\[v e^t + 2v' e^t + v'' e^t - 2(v e^t + v' e^t) + v e^t = 0\]\[v e^t + 2v' e^t + v'' e^t - 2v e^t - 2v' e^t + v e^t = 0\]\[v'' = 0\]

\(v = at + b\) (we can use ANY \(a, b\) (\(a \neq 0\)) as \(v\) for \(y_2 = v y_1\))

let's use \(v = t\)

then \(y_1 = e^t\) and \(y_2 = t e^t\)

\[W[y_1, y_2] = \begin{vmatrix} e^t & t e^t \\ e^t & t e^t + e^t \end{vmatrix}\]\[= t e^{2t} + e^{2t} - t e^{2t} = e^{2t} \neq 0 \text{ (good!)}\]
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if the differential eq. has constant coefficient

\[ y_2 = t y_1 \text{ is ALWAYS a valid 2nd solution.} \]

but Reduction of Order can be used to find missing solution even for non-constant coefficients

\[ t^2 y'' + t y' - 9y = 0 \quad \text{(this is an example of Euler's equation)} \]

has one solution \( y_1 = t^3 \) (somehow we found this)

\[ \begin{aligned} y_2 &= ? \\ y_2 &= v y_1 = v t^3 \\ y_2' &= 3 v t^2 + v' t^3 \\ y_2'' &= 6 v t + 6 v' t^2 + v'' t^3 \end{aligned} \]
\[ t^2 (6 v t + 6 v' t^2 + v'' t^3) + t (3 v t^2 + v' t^3) - 9 (v t^3) = 0 \]
\[ \vdots \]
\[ t^5 v'' + 7 v' t^4 = 0 \]
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\[ v'' = -7 v' t^{-1} \quad \quad v'' = \frac{d(v')}{dt} \]
\[ \frac{d(v')}{dt} = -\frac{7 v'}{t} \quad \text{separable in } v' \text{ and } t \]

solve for \( v' \dots \) \( v' = c t^{-7} \)

then \( v = -\frac{c}{6} t^{-6} + d \)

choose \( c = -6, d = 0 \)

\( v = t^{-6} \)

so \( y_2 = v y_1 = t^{-6} \cdot t^3 = t^{-3} \)