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7.2 (continued)

use Laplace transform for linear system

\[ \begin{aligned} x' &= x + 2y \\ y' &= 2x + y \end{aligned} \quad x(0)=1, y(0)=0 \]

transform each equation

\[ \mathcal{L}\{x\} = X \quad \mathcal{L}\{y\} = Y \]

\[ \mathcal{L}\{x'\} = \mathcal{L}\{x + 2y\} \]

\[ \begin{aligned} sX - x(0) &= X + 2Y \\ sY - y(0) &= 2X + Y \end{aligned} \]

\[ sX - X - 2Y = 1 \]

\[ \left. \begin{aligned} (s-1)X - 2Y &= 1 \\ -2X + (s-1)Y &= 0 \end{aligned} \right\} \begin{bmatrix} s-1 & -2 \\ -2 & s-1 \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

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the system can be solved in many ways

Cramer's rule is particularly useful

\[ \underline{X} = \frac{\begin{vmatrix} 1 & -2 \\ 0 & s-1 \end{vmatrix}}{\begin{vmatrix} s-1 & -2 \\ -2 & s-1 \end{vmatrix}} \]

det. of coeff. matrix 1st column replaced by the right side

determinant of coeff. matrix

\[ \underline{X} = \frac{s-1}{(s-1)^2 - 4} = \frac{s-1}{s^2 - 2s - 3} = \frac{s-1}{(s-3)(s+1)} = \dots \]\[ = \frac{1/2}{s-3} + \frac{1/2}{s+1} \]\[ x(t) = \frac{1}{2} e^{3t} + \frac{1}{2} e^{-t} \]

\[ \underline{Y} = \frac{\begin{vmatrix} s-1 & 1 \\ -2 & 0 \end{vmatrix}}{\begin{vmatrix} s-1 & -2 \\ -2 & s-1 \end{vmatrix}} = \frac{2}{(s-3)(s+1)} = \dots \quad y(t) = \frac{1}{2} e^{3t} - \frac{1}{2} e^{-t} \]

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7.3 Translation of Laplace Transform

we know \[ \mathcal{L} \{ f(t) \} = F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]

what happens to \( f(t) \) if we do a translation in \( s \) domain

\( F(s-a) \rightarrow f(t) \text{ ?} \)
shift to the right by \( a \)

\[ F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]

\[ F(s-a) = \int_{0}^{\infty} f(t) e^{-(s-a)t} dt \]

\[ F(s-a) = \int_{0}^{\infty} [f(t) e^{at}] e^{-st} dt = \mathcal{L} \{ f(t) e^{at} \} \]

Translation Theorem (1)

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from table : \(\mathcal{L}\{1\} = \frac{1}{s} = F(s)\)

\(F(s-a) = \frac{1}{s-a} = \mathcal{L}\{1 \cdot e^{at}\} = \mathcal{L}\{e^{at}\}\)

\(\mathcal{L}\{\sin(at)\} = \frac{a}{s^2+a^2} = F(s)\)

\(F(s-c) = \frac{a}{(s-c)^2+a^2} = \mathcal{L}\{e^{ct} \sin(at)\}\)

Example

\[ y'' - 2y' + 5y = 8e^t \]

\[ y(0) = 2, y'(0) = 4 \]

\[ s^2 Y - s y(0) - y'(0) - 2(s Y - y(0)) + 5Y = \frac{8}{s-1} \]

\[ (s^2 - 2s + 5)Y = 2s + 4 - 4 + \frac{8}{s-1} \]

\[ Y = \frac{2s}{s^2 - 2s + 5} + \frac{8}{(s-1)(s^2 - 2s + 5)} \]

\[ = \frac{2s(s-1) + 8}{(s-1)(s^2 - 2s + 5)} \]

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\[ Y = \frac{2s^2 - 2s + 8}{(s-1)[(s-1)^2 + 4]} \]\[ = \frac{A}{s-1} + \frac{Bs + C}{(s-1)^2 + 4} \]\[ 2s^2 - 2s + 8 = A[(s-1)^2 + 4] + (Bs + C)(s-1) \]\[ = A(s^2 - 2s + 1) + 4A + Bs^2 - Bs + Cs - C \]\[ 2s^2 - 2s + 8 = (A+B)s^2 + (-2A - B + C)s + (5A - C) \]

\( A + B = 2 \)
\( -2A - B + C = -2 \)
\( 5A - C = 8 \)

\( 4A = 8 \quad A = 2, B = 0, C = 2 \)

\( Y = \frac{2}{s-1} + \frac{2}{(s-1)^2 + 4} \)

looks like \( \frac{2}{s^2 + 4} \) shift by 1

\( e^t \sin(2t) \)

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ℳ: \( y = 2e^t + e^t \sin(2t) \)

1st Translation Theorem: \( F(s-a) = \mathcal{L} \{ e^{at} f(t) \} \)

2nd Translation Theorem: \( \mathcal{L} \{ f(t-a) \} = ? \)

\[ \mathcal{L} \{ f(t-a) \} = \int_{0}^{\infty} f(t-a) e^{-st} dt \]

Figure: Graph of a function f(t) on a coordinate system with axes f(t) and t. The curve starts at the origin.

Figure: Graph of a shifted function f(t) on a coordinate system with axes f(t) and t. The curve starts at a point labeled 'a' on the t-axis.

= 0 for \( t < a \)

let \( \tau = t - a \)

\[ = \int_{a}^{\infty} f(t-a) e^{-st} dt \]

delay in time