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7.2 (continued)

can use Laplace transform to solve systems

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

basic idea: transform, solve in s-domain, then inverse transform

\( \mathcal{L}\{x\} = X \quad \mathcal{L}\{y\} = Y \)

\( \mathcal{L}\{x'\} = \mathcal{L}\{x + 2y\} \)

\( sX - x(0) = X + 2Y \)

\( sY - y(0) = 2X + Y \)

rewrite

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

solve for \( X, Y \)

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\[ \begin{bmatrix} s-1 & -2 \\ -2 & s-1 \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

solve by row reduction, multiply by inverse of \( \begin{bmatrix} s-1 & -2 \\ -2 & s-1 \end{bmatrix} \) or Cramer's rule.

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

determinant of coefficient matrix w/ 1st column replaced by right side ( because \( X \) is the 1st variable )

determinant of coefficient matrix

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

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

7.3 Translation of Laplace Transform

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

Shift horizontally by \( a \) in \( s \)-domain: \( F(s-a) \)

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

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

multiplication of \( e^{at} \) in \( t \)-domain is a shift by \( a \) in \( s \)-domain.

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

shift in \(s\) by \(a \rightarrow\) multiplication by \(e^{at}\) in \(t\)

\(F = \frac{1}{s-a} \rightarrow f = 1 \cdot e^{at} = e^{at}\) which agrees with table entries

same with \(\sin(at) \rightarrow \frac{a}{s^2+a}\)

\[ \frac{a}{(s-c)^2+a} \rightarrow e^{ct} \sin(at) \]

example \(y'' - 2y' + 5y = 8e^t\)

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

\[ s^2 Y - s y(0) - y'(0) - 2 (sY - y(0)) + 5Y = \frac{8}{s-1} \]\[ (s^2 - 2s + 5) Y = 2s + 4 - 4 + \frac{8}{s-1} \]\[ = 2s + \frac{8}{s-1} = \frac{2s^2 - 2s + 8}{s-1} \]

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

if cannot be factored, put into \( (s-a)^2 \pm b^2 \)

\[ s^2 - 2s + 5 = s^2 - 2s + 1 + 4 \]\[ = (s-1)^2 + 4 \]

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

\[ \begin{matrix} A + B = 2 \\ -2A + C - B = -2 \\ 5A - C = 8 \end{matrix} \quad \dots \quad \begin{matrix} A = 2 \\ B = 0 \\ C = 2 \end{matrix} \]

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

\( \frac{2}{s^2 + 4} \) shifted by 1
\( \sin(2t) \) times \( e^t \)

\[ y = 2e^t + e^t \sin(2t) \]

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

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

assuming \( f(t-a) = 0 \) for \( t < a \)