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7.2 Laplace Transform of Initial-Value Problems

solve something like \( y'' + 4y = 8 \), \( y(0) = 0 \), \( y'(0) = 6 \)

For \( y'' + 4y = 8 \):
- mass = 1
- spring constant = 4
- force of (upward) 8
For \( y(0) = 0 \): initial displacement
For \( y'(0) = 6 \): initial velocity

basic idea: Laplace transform of both sides

\( \mathcal{L} \{ y(t) \} = Y(s) \)
\( \mathcal{L} \{ y'(t) \} = ? \)
\( \mathcal{L} \{ y''(t) \} = ? \)

\[ \mathcal{L} \{ y' \} = \int_{0}^{\infty} y' e^{-st} dt = \lim_{a \to \infty} \int_{0}^{a} y' e^{-st} dt \]

\( u = e^{-st} \quad dv = y' dt \)

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

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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: \( y(a) e^{-sa} \to 0 \) for \( s > 0 \), and \( \int_0^\infty y e^{-st} dt = \mathcal{L}\{y\} = Y \).

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

likewise,

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

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

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

Laplace transform: \( \mathcal{L}\{y''\} + \mathcal{L}\{4y\} = \mathcal{L}\{8\} \)

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

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

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solve for \( Y \): \( (s^2 + 4)Y = 6 + \frac{8}{s} \)

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

problem "solved" but we would like to have \( y(t) \)

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

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

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

\[ \mathcal{L}^{-1} \left\{ \frac{6}{s^2 + 4} \right\} = 6 \mathcal{L}^{-1} \left\{ \frac{1}{s^2 + 2^2} \right\} \]

want a 2 here

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

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

table entries are all \( \frac{1}{s-a} \) or \( \frac{1}{s^2 \pm a^2} \) no \( \frac{1}{s(s^2 \pm a^2)} \)

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partial fraction expansion: \[ \frac{8}{s(s^2+4)} = \frac{A}{s} + \frac{Bs+C}{s^2+4} \]

\(Bs+C\) — linear (one degree lower than denom)
\(s^2+4\) — quadratic

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

\(A+B = 0\)
\(C = 0\)
\(4A = 8\) so \(A=2\), \(B=-2\)

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

so, \(y(t) = 3 \sin(2t) + 2 - 2 \cos(2t)\)

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In some cases, there is an alternative to doing partial fraction expansion

we know \(\mathcal{L} \{ f'(t) \} = s F(s) - f(0)\)

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

let \(g(t) = \int_0^t f(\tau) d\tau\)

then from calculus,

\[ g'(t) = f(t) \]

\[ \mathcal{L} \{ g'(t) \} = s G - \underbrace{g(0)}_{0} = s G \]

\[ \mathcal{L} \{ f(t) \} = s G = s \mathcal{L} \left\{ \int_0^t f(\tau) d\tau \right\} \]

or \[ F = s \mathcal{L} \left\{ \int_0^t f(\tau) d\tau \right\} \]

or \[ \frac{F}{s} = \mathcal{L} \left\{ \int_0^t f(\tau) d\tau \right\} \rightarrow \]

\[ \mathcal{L}^{-1} \left\{ \frac{F}{s} \right\} = \int_0^t f(\tau) d\tau \]

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

find \( \mathcal{L}^{-1} \left\{ \frac{8}{s^2+4} \right\} \) then integrate

\( 4 \sin(2t) \)

\[ = \int_{0}^{t} 4 \sin(2\tau) d\tau = -2 \cos(2\tau) \Big|_0^t = -2 \cos(2t) + 2 \]