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3.2 Solutions of Linear Homogeneous Eqs

If \[ y'' + p(t)y' + q(t)y = 0 \] we know there are two solutions \( y_1 = e^{r_1 t} \) and \( y_2 = e^{r_2 t} \) where \( r_1 \) and \( r_2 \) are roots of the characteristic eq.

The linear combination \[ y = c_1 y_1 + c_2 y_2 \] is also a solution, called the general solution.

Why?

If \( y_1 \) and \( y_2 \) are solutions, then

\[ y_1'' + p(t)y_1' + q(t)y_1 = 0 \]

\[ y_2'' + p(t)y_2' + q(t)y_2 = 0 \]

If \( y = c_1 y_1 + c_2 y_2 \) is a solution for any \( c_1, c_2 \), then it also satisfies the DE.

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\[ y'' + p(t)y' + q(t)y = 0 \]

\[ y' = c_1 y_1' + c_2 y_2' \]

\[ y'' = c_1 y_1'' + c_2 y_2'' \]

\[ c_1 y_1'' + c_2 y_2'' + p(t) [c_1 y_1' + c_2 y_2'] + q(t) [c_1 y_1 + c_2 y_2] = 0 \]

\[ c_1 [y_1'' + p(t)y_1' + q(t)y_1] + c_2 [y_2'' + p(t)y_2' + q(t)y_2] = 0 \]

zero, because this is the DE and \( y_1 \) is solution

zero, same reason

→ this proves that \( y = c_1 y_1 + c_2 y_2 \) is a solution for ANY \( c_1, c_2 \).

BUT, we only care about \( c_1 \) and \( c_2 \) that satisfy the IC's : \[ y(t_0) = y_0, \quad y'(t_0) = y_0' \]

can we always find \( c_1 \) and \( c_2 \) for ANY IC's?

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Given \[ y'' + p(t)y' + g(t)y = 0 \] has solutions \( y_1, y_2 \) and general solution \( y = c_1 y_1 + c_2 y_2 \) and IC's: \( y(t_0) = y_0, \) \( y'(t_0) = y'_0 \)

to find \( c_1, c_2 \), we solve

\[ y_0 = c_1 y_1(t_0) + c_2 y_2(t_0) \]\[ y'_0 = c_1 y'_1(t_0) + c_2 y'_2(t_0) \]

rewrite:

\[ \begin{bmatrix} y_1(t_0) & y_2(t_0) \\ y'_1(t_0) & y'_2(t_0) \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} y_0 \\ y'_0 \end{bmatrix} \]\[ \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} y_1(t_0) & y_2(t_0) \\ y'_1(t_0) & y'_2(t_0) \end{bmatrix}^{-1} \begin{bmatrix} y_0 \\ y'_0 \end{bmatrix} \]

the inverse exists if and only if

\[ \det \left( \begin{bmatrix} y_1(t_0) & y_2(t_0) \\ y'_1(t_0) & y'_2(t_0) \end{bmatrix} \right) \neq 0 \]

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The Wronskian

this determinant is called the Wronskian

\[ W(y_1, y_2)(t) = \begin{vmatrix} y_1(t) & y_2(t) \\ y'_1(t) & y'_2(t) \end{vmatrix} \]

\( c_1 \) and \( c_2 \) also always exist for any IC's if \( W(y_1, y_2)(t_0) \neq 0 \)

Example

\( y_1 = \cos t \quad y_2 = \sin t \)

\[ W(y_1, y_2) = \begin{vmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{vmatrix} = (\cos t)^2 - (-\sin t)(\sin t) \]\[ = \cos^2 t + \sin^2 t = 1 \]

never zero for ANY \( t_0 \)

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Example: Finding a Function from the Wronskian

If the Wronskian of \( f \) and \( g \) is \( 3e^{4t} \) and \( f(t) = e^{2t} \), what is \( g(t) \)?

\[ W = 3e^{4t} = \begin{vmatrix} e^{2t} & g \\ 2e^{2t} & g' \end{vmatrix} \]

\[ 3e^{4t} = e^{2t}g' - 2e^{2t}g \]

1st order linear

\[ e^{2t}g' - 2e^{2t}g = 3e^{4t} \]

Dividing by \( e^{2t} \):

\[ g' - 2g = 3e^{2t} \]

Integrating factor: \( \mu = e^{-2t} \)

\[ e^{-2t}g' - 2e^{-2t}g = 3 \]

\[ \frac{d}{dt}(e^{-2t}g) = 3 \]

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\[ e^{-2t}g = 3t + C \]

\[ g = 3te^{2t} + Ce^{2t} = (3t + C)e^{2t} \]

Note on Wronskian Order

Placement of \( f \) and \( g \) is irrelevant;

\[ \begin{vmatrix} f & g \\ f' & g' \end{vmatrix} \text{ and } \begin{vmatrix} g & f \\ g' & f' \end{vmatrix} \text{ are both ok.} \]

Fundamental Solutions

If \( W(y_1, y_2)(t_0) \neq 0 \), then \( y_1 \) and \( y_2 \) are the fundamental solutions and they form the fundamental set of solutions.

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Existence and Uniqueness of solutions

The initial-value problem

\[ y'' + p(t)y' + q(t)y = g(t) \]

(ANY Linear but not necessarily homogeneous)

\[ y(t_0) = y_0, \quad y'(t_0) = y'_0 \]

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

Example

\[ 2t^2 y'' + 3t y' - y = 0, \quad y(1) = 3 \]\[ y'' + \frac{3}{2t} y' - \frac{1}{2t^2} y = 0 \]

\( p \)

\( \frac{3}{2t} \)

\( q \)

\( -\frac{1}{2t^2} \)

\( g \)

\( 0 \)

\( p, q, g \) continuous on \( (-\infty, 0) \) and \( (0, \infty) \)

initial \( t \) is 1, so interval is \( (0, \infty) \)

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Quick review of some complex numbers (for 3.3)

\[ i^2 = -1 \quad i = \pm\sqrt{-1} \]

imaginary number

\[ a + bi \]

complex number (\( a, b \) are constants)

\[ 3 + 2i \]

3 is the real part

2 is the imaginary part

Euler's Identity

complex exponential:

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

\[ e^{(3+2i)t} = e^{3t} e^{i(2t)} = e^{3t} (\cos 2t + i \sin 2t) \]