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3.1 (continued)

\[ ay'' + by' + cy = 0 \quad y(t_0) = y_0, \ y'(t_0) = y'_0 \]

solutions: \( y = e^{rt} \) \( r \): constant

Sub into the diff. eq.

\[ ar^2 e^{rt} + br e^{rt} + c e^{rt} = 0 \] \[ e^{rt} (ar^2 + br + c) = 0 \quad e^{rt} \neq 0 \]
\[ ar^2 + br + c = 0 \]
characteristic eq.

notice the structure compared to

\[ ay'' + by' + cy = 0 \]

two solutions: \( y_1 = e^{r_1 t} \) , \( y_2 = e^{r_2 t} \) \( r_1, r_2 \) are roots of char. eq.

general solution: \( y = c_1 e^{r_1 t} + c_2 e^{r_2 t} \)

\( c_1, c_2 \) come from initial conditions

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example

\[ y'' + 8y' - 9y = 0 \quad y(0) = 1, \ y'(0) = 0 \]

char. eq: \( r^2 + 8r - 9 = 0 \)

\( (r + 9)(r - 1) = 0 \)

\( r = -9, \ r = 1 \)

general solution: \( y = c_1 e^{-9t} + c_2 e^t \)

find \( c_1, c_2 \) from \( y(0) = 1, \ y'(0) = 0 \)

\[ y(0) = 1 \rightarrow 1 = c_1 + c_2 \]

to use \( y'(0) = 0 \), must have \( y' \)

from \( y = c_1 e^{-9t} + c_2 e^t \) we get

\[ y' = -9c_1 e^{-9t} + c_2 e^t \]

cannot call this \( c_1 \) since we care about value of \( c_1 \)

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Solving for Constants

\[ y'(0) = 0 \rightarrow \left. \begin{aligned} 0 &= -9C_1 + C_2 \\ 1 &= C_1 + C_2 \end{aligned} \right\} \text{ solve simultaneously} \]

Multiply the second equation by 9:

\[ 9 = 9C_1 + 9C_2 \]

Add to the first equation:

\[ 0 = -9C_1 + C_2 \] \[ 9 = 10C_2 \implies C_2 = \frac{9}{10} \]

Then solve for \( C_1 \):

\[ C_1 = 1 - C_2 = \frac{1}{10} \]

Particular Solution

\[ y = \frac{1}{10}e^{-9t} + \frac{9}{10}e^t \]

\[ y' = -\frac{9}{10}e^{-9t} + \frac{9}{10}e^t \]

Sketch

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A coordinate graph showing y versus t. The curve starts at y=1 on the y-axis with a horizontal slope and curves upward.

\[ y(t_0) = y_0 \] : initial position

\[ y'(t_0) = y_0' \] : initial slope

\[ y(0) = 1 \]

\[ y'(0) = 0 \]

Slope of curve at \( t = 0 \) (flat at \( t = 0 \))

\[ y' = -\frac{9}{10}e^{-9t} + \frac{9}{10}e^t \]

\( < e^t \) as \( t \) increases

\[ 0 \le \text{slope} < \frac{9}{10} \]

as \( t \to \infty \)

\[ \lim_{t \to \infty} e^{-9t} = 0 \]

\[ \lim_{t \to \infty} y' = \lim_{t \to \infty} \frac{9}{10}e^t \]

\[ \lim_{t \to \infty} y = \frac{9}{10}e^t \]

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Example: Second-Order Linear Homogeneous Differential Equation

\[ y'' + 5y' + 6y = 0, \quad y(0) = 2, \quad y'(0) = \beta \quad (\beta > 0) \]

initial slope is pos.

A sketch of a graph on y and t axes starting at y=2 and increasing initially.

Characteristic Equation

\[ r^2 + 5r + 6 = 0 \] \[ (r + 3)(r + 2) = 0 \] \[ r = -3, \quad r = -2 \]

General Solution and Derivatives

\[ y = C_1 e^{-3t} + C_2 e^{-2t} \] \[ y' = -3C_1 e^{-3t} - 2C_2 e^{-2t} \]

Applying Initial Conditions

  • \( y(0) = 2 \rightarrow 2 = C_1 + C_2 \)
  • \( y'(0) = \beta \rightarrow \beta = -3C_1 - 2C_2 \)

Asymptotic Behavior

Since \( \lim_{t \to \infty} e^{-3t} = \lim_{t \to \infty} e^{-2t} = 0 \), then:

\[ \lim_{t \to \infty} y = 0 \]

\( y'(0) > 0 \), so initial go up, reach a max, then approach 0.

A graph showing a curve starting above the t-axis, rising to a peak, and then decaying toward zero.
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Solving the System

after solving the system,

\[ C_1 = -(4 + \beta), \quad C_2 = \beta + 6 \] \[ y = -(4 + \beta)e^{-3t} + (\beta + 6)e^{-2t} \]

Finding Maximum Value

time to reach max \( y \)?

Graph of y vs t showing a peak at point (tm, ym) before decaying.
\[ y' = 3(4 + \beta)e^{-3t} - 2(\beta + 6)e^{-2t} \] \[ \text{max } y \rightarrow y' = 0 \] \[ 3(4 + \beta)e^{-3t} - 2(\beta + 6)e^{-2t} = 0 \] \[ 3(4 + \beta)e^{-3t} = 2(\beta + 6)e^{-2t} \] \[ 3(4 + \beta) = 2(\beta + 6) e^t \] \[ e^t = \frac{3(4 + \beta)}{2(\beta + 6)} \]
\[ t_m = \ln\left( \frac{12 + 3\beta}{2\beta + 12} \right) \]

note: \( \lim_{\beta \to \infty} t_m = \ln\left(\frac{3}{2}\right) \)

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Sub \( t_m \) into \( y \) to find \( y_m \)

after some algebra,

\[ y_m = \frac{4}{27} \frac{(6+\beta)^3}{(4+\beta)^2} \]
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3.2 Fundamental Solutions of Linear Homogeneous Eqs.

Solutions of \( ay'' + by' + cy = 0 \), \( y(t_0) = y_0 \), \( y'(t_0) = y'_0 \) are always unique

but what about \( y'' + p(t)y' + q(t)y = g(t) \)

very similar to \( y' + p(t)y = g(t) \), \( y(t_0) = y_0 \)

when \( p, g \) are both continuous and containing \( y(t_0) = y_0 \)

2nd order: same idea

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

find interval of \( t \) where all are continuous and containing \( t_0 \)