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7.2 Laplace Transform of Initial-Value Problems

Solve, for example, mass-spring problem like

\[ y'' + 4y = 8 \quad y(0) = 0, y'(0) = 6 \]

mass = 1

spring constant = 4

external (upward) force = 8

initial displacement = 0

initial velocity (up) = 6

basic idea: transform both sides, solve for \( \mathcal{L}\{y\} = Y \)

then inverse transform to find \( y(t) \)

\( \mathcal{L}\{y\} = Y \)

\( \mathcal{L}\{y'\} = ? \)

\[ \mathcal{L}\{y'\} = \int_0^\infty y' e^{-st} dt = \lim_{a \to \infty} \int_0^a y' e^{-st} dt \]

\( u = e^{-st} \)

\( du = -s e^{-st} dt \)

\( dv = y' dt \)

\( v = y \)

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\[ = \lim_{a \to \infty} \left( y(t)e^{-st} \Big|_0^a + s \int_0^a y e^{-st} dt \right) \]

\[ = \lim_{a \to \infty} (y(a)e^{-sa} - y(0)) + s \int_0^\infty y e^{-st} dt \]

(Note: \( y(a)e^{-sa} \) goes to 0 so \( s > 0 \); the integral \( \int_0^\infty y e^{-st} dt \) is \( \mathcal{L}\{y\} = Y \))

\[ \mathcal{L}\{y'\} = sY - y(0) \]

\( s > 0 \)

\[ \mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0) \]

\[ \mathcal{L}\{y^{(n)}\} = s^n Y - s^{n-1} y(0) - s^{n-2} y'(0) - \dots - y^{(n-1)}(0) \]

back to \( y'' + 4y = 8 \quad y(0) = 0, y'(0) = 6 \)

\[ \mathcal{L}\{y''\} + \mathcal{L}\{4y\} = \mathcal{L}\{8\} \]

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

(Substituting initial conditions: \( sy(0) = 0 \) and \( y'(0) = 6 \))

solve for \( Y \)

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\[ (s^2+4)Y = 6 + \frac{8}{s} \]\[ Y = \frac{6}{s^2+4} + \frac{8}{s(s^2+4)} \]

solution in s-domain
want solution in t-domain

\[ y(t) = \mathcal{L}^{-1} \left\{ \frac{6}{s^2+4} \right\} + \mathcal{L}^{-1} \left\{ \frac{8}{s(s^2+4)} \right\} \]

directly from table: \( \mathcal{L}^{-1} \left\{ \frac{a}{s^2+a^2} \right\} = \sin(at) \)

\[ \mathcal{L}^{-1} \left\{ \frac{6}{s^2+2^2} \right\} = \mathcal{L}^{-1} \left\{ 3 \cdot \frac{2}{s^2+2^2} \right\} = 3 \sin(2t) \]

\( \mathcal{L}^{-1} \left\{ \frac{8}{s(s^2+4)} \right\} \) is a bit more complicated

no \( \frac{1}{s(s^2+a^2)} \) on the table, but there is \( \frac{1}{s}, \frac{1}{s^2+a^2} \)

nice if \[ \frac{8}{s(s^2+4)} = \frac{A}{s} + \frac{Bs+C}{s^2+4} \]

partial fraction expansion
numerator is one degree lower than denominator

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\[ 8 = A(s^2+4) + (Bs+C)s \]

\[ 0s^2+0s+8 = (A+B)s^2 + Cs + 4A \]

\[ A+B=0 \]

\[ C=0 \]

\[ 4A=8 \quad \text{so} \quad A=2, B=-2 \]

\[ \mathcal{L}^{-1} \left\{ \frac{8}{s(s^2+4)} \right\} = \mathcal{L}^{-1} \left\{ \frac{2}{s} - \frac{2s}{s^2+4} \right\} \]

\[ = 2 - 2 \cos(2t) \]

so,

\[ y(t) = 3 \sin(2t) + 2 - 2 \cos(2t) \]

another transform to look at:

we know \( \mathcal{L} \{ y' \} = sY - y(0) \)

\[ \mathcal{L} \{ f'(t) \} = sF - f(0) \]

what is \( \mathcal{L} \left\{ \int_0^t f(\tau) d\tau \right\} ? \)

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let \( g(t) = \int_{0}^{t} f(\tau) d\tau \)

then, from calculus, \( g'(t) = f(t) \)

\[ \mathcal{L} \{ g'(t) \} = sG - \underbrace{g(0)}_{0} = \mathcal{L} \{ f(t) \} = F \]

\[ sG = F \quad G = \mathcal{L} \{ g(t) \} = \mathcal{L} \left\{ \int_{0}^{t} f(\tau) d\tau \right\} \]

so, \( s \mathcal{L} \left\{ \int_{0}^{t} f(\tau) d\tau \right\} = F \)

\[ \mathcal{L} \left\{ \int_{0}^{t} f(\tau) d\tau \right\} = \frac{F}{s} \leftrightarrow \int_{0}^{t} f(\tau) d\tau = \mathcal{L}^{-1} \left\{ \frac{F}{s} \right\} \]

for example, \( \mathcal{L}^{-1} \left\{ \frac{8}{s(s^2+4)} \right\} = \mathcal{L}^{-1} \left\{ \frac{\frac{8}{s^2+4}}{s} \right\} \)

\( F = \frac{8}{s^2+4} \) find \( f \) then \( \int_{0}^{t} f d\tau \)

\[ \mathcal{L}^{-1} \left\{ \frac{8}{s^2+4} \right\} = 4 \sin(2t) \]

\[ \int_{0}^{t} 4 \sin(2\tau) d\tau = \dots = -2 \cos(2t) + 2 \]

same as from partial fraction