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Review: Existence of Laplace Transforms
From Section 7.1, if a function \(f(t)\) is piecewise continuous for \(t \ge 0\) and is of exponential order as \(t \to \infty\) (there exist nonnegative constants \(M\), \(c\), and \(T\) such that \(|f(t)| \le Me^{ct}\) for \(t \ge T\)), then its Laplace transform \(F(s)\) exists.
We apply Laplace transformations to solve a linear differential equation with constant coefficients:
\[ a x''(t) + b x'(t) + c x(t) = f(t), \quad x(0) = x_0, \quad x'(0) = x'_0 \]
By the linearity of the Laplace transformation, we can transform each term in the equation:
\[ a\mathcal{L}\{x''(t)\} + b\mathcal{L}\{x'(t)\} + c\mathcal{L}\{x(t)\} = \mathcal{L}\{f(t)\} \]
The key to this method is expressing the transforms of derivatives in terms of the transform of the function itself.
Theorem 1: Transforms of Derivatives
Suppose \(f(t)\) is continuous and piecewise smooth for \(t \ge 0\) and is of exponential order as \(t \to \infty\). Then \(\mathcal{L}\{f'(t)\}\) exists for \(s > c\), and:
\[ \mathcal{L}\{f'(t)\} = sF(s) - f(0) \]
(I) Solutions of Initial Value Problems
Using Theorem 1 repeatedly, we can find the transforms of higher-order derivatives:
\(
\begin{aligned}
\mathcal{L}\{x'(t)\} &= sX(s) - x(0) \\
\mathcal{L}\{x''(t)\} &= s\mathcal{L}\{x'(t)\} - x'(0) \\
&= s(sX(s) - x(0)) - x'(0) \\
&= s^2X(s) - sx(0) - x'(0)
\end{aligned}
\)
Corollary: Transform of Higher Derivatives
Suppose that \(f, f', \dots, f^{(n-1)}\) are continuous and piecewise smooth for \(t \ge 0\) and are of exponential order. Then:
\[ \mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \dots - f^{(n-1)}(0) \]
Example 1
Solve the Initial Value Problem: \( x'' - x' - 6x = 0; \quad x(0) = 2, \quad x'(0) = -1 \).
\(
\begin{aligned}
\mathcal{L}\{x'\} &= sX(s) - 2 \\
\mathcal{L}\{x''\} &= s^2X(s) - 2s + 1
\end{aligned}
\)
Substitute into the differential equation:
\(
\begin{aligned}
(s^2X(s) - 2s + 1) - (sX(s) - 2) - 6X(s) &= 0 \\
(s^2 - s - 6)X(s) - 2s + 3 &= 0 \\
(s^2 - s - 6)X(s) &= 2s - 3 \\
X(s) &= \frac{2s - 3}{(s+2)(s-3)}
\end{aligned}
\)
Partial fraction decomposition (linear factors case):
\(
\begin{aligned}
\frac{2s - 3}{(s+2)(s-3)} &= \frac{A}{s-3} + \frac{B}{s+2} \\
2s - 3 &= A(s+2) + B(s-3)
\end{aligned}
\)
If \(s = 3 \implies 3 = 5A \implies A = 3/5\).
\[ X(s) = \frac{3}{5}\left(\frac{1}{s-3}\right) + \frac{7}{5}\left(\frac{1}{s+2}\right) \implies x(t) = \mathcal{L}^{-1}\{X(s)\} = \frac{3}{5}e^{3t} + \frac{7}{5}e^{-2t} \]
Example 2
Solve the Initial Value Problem: \( x'' + 4x = \cos t; \quad x(0) = 0, \quad x'(0) = 0 \).
Transformed equation:
\(
\begin{aligned}
s^2X(s) + 4X(s) &= \frac{s}{s^2 + 1} \\
X(s) &= \frac{s}{(s^2 + 4)(s^2 + 1)}
\end{aligned}
\)
Partial fraction decomposition (irreducible quadratic case):
\(
\begin{aligned}
\frac{s}{(s^2 + 4)(s^2 + 1)} &= \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 4} \\
s &= (As + B)(s^2 + 4) + (Cs + D)(s^2 + 1)
\end{aligned}
\)
Equating coefficients:
\(s^3: A + C = 0\)
\(s^2: B + D = 0\)
\(s^1: 4A + C = 1\)
\(s^0: 4B + D = 0\)
Solving yields \(A = 1/3\), \(C = -1/3\), and \(B = D = 0\).
\[ X(s) = \frac{1}{3}\left(\frac{s}{s^2 + 1}\right) - \frac{1}{3}\left(\frac{s}{s^2 + 4}\right) \implies x(t) = \frac{1}{3}\cos t - \frac{1}{3}\cos 2t \]
How to use Laplace Transform to solve an I.V.P.
Differential Equation in \(x(t)\) Algebraic Equation in \(X(s)\) \(X(s)\) isolated Solution \(x(t)\) of the Differential Equation
(II) Solve Linear Systems
Example 3
Solve algebraically:
\(
\begin{cases}
2x'' = -6x + 2y \\
y'' = 2x - 2y + 40\sin(3t)
\end{cases} \quad \text{with } x(0) = x'(0) = y(0) = y'(0) = 0
\)
Write \(X(s) = \mathcal{L}\{x(t)\}\) and \(Y(s) = \mathcal{L}\{y(t)\}\). The transformed equations are:
\(
\begin{cases}
2s^2 X(s) = -6X(s) + 2Y(s) \implies (s^2 + 3)X(s) - Y(s) = 0 \\
s^2 Y(s) = 2X(s) - 2Y(s) + \frac{120}{s^2 + 9} \implies -2X(s) + (s^2 + 2)Y(s) = \frac{120}{s^2 + 9}
\end{cases}
\)
By Cramer's Rule or the Method of Elimination:
\(
\begin{aligned}
X(s) &= \frac{120}{(s^2 + 1)(s^2 + 4)(s^2 + 9)} = \frac{5}{s^2 + 1} - \frac{8}{s^2 + 4} + \frac{3}{s^2 + 9} \\
Y(s) &= \frac{120(s^2 + 3)}{(s^2 + 1)(s^2 + 4)(s^2 + 9)} = \frac{10}{s^2 + 1} + \frac{8}{s^2 + 4} - \frac{18}{s^2 + 9}
\end{aligned}
\)
Conclusion:\[\begin{cases} x(t) = 5\sin t - 4\sin 2t + \sin 3t \\ y(t) = 10\sin t + 4\sin 2t - 6\sin 3t \end{cases} \]
(III) Transformation of Integrals
It's reasonable to expect the inverse operation of integration to correspond to division of the transform by \(s\).
Theorem 2: Transform of Integrals
If \(f(t)\) is a piecewise continuous function for \(t \ge 0\) and satisfies the condition of exponential order \(|f(t)| \le Me^{ct}\) for \(t \ge T\), then:
\[ \mathcal{L}\left\{ \int_{0}^{t} f(\tau) d\tau \right\} = \frac{F(s)}{s} \quad \text{for } s > c \]
Equivalently:
\[ \mathcal{L}^{-1}\left\{ \frac{F(s)}{s} \right\} = \int_{0}^{t} f(\tau) d\tau = \int_{0}^{t} \mathcal{L}^{-1}\{F(s)\} d\tau \]
Example 4
Find \(\mathcal{L}^{-1}\left\{ \frac{1}{s^2(s-a)} \right\}\).
Applying the theorem once:
\(
\begin{aligned}
\mathcal{L}^{-1}\left\{ \frac{1}{s(s-a)} \right\} &= \int_{0}^{t} \mathcal{L}^{-1}\left\{ \frac{1}{s-a} \right\} d\tau \\
&= \int_{0}^{t} e^{a\tau} d\tau \\
&= \frac{1}{a}(e^{at} - 1)
\end{aligned}
\)
Applying the theorem a second time for the remaining \(1/s\):
\(
\begin{aligned}
\mathcal{L}^{-1}\left\{ \frac{1}{s^2(s-a)} \right\} &= \frac{1}{a} \int_{0}^{t} (e^{a\tau} - 1) d\tau \\
&= \frac{1}{a} \left[ \frac{1}{a}e^{a\tau} - \tau \right]_{0}^{t}
\end{aligned}
\)
Conclusion:\[\mathcal{L}^{-1}\left\{ \frac{1}{s^2(s-a)} \right\} = \frac{1}{a^2}(e^{at} - at - 1) \]
Example 5
Find the inverse transform of \( F(s) = \frac{1}{s^2(s^2 + 1)} \).
Method 1: Use Theorem 2.
\(
\begin{aligned}
\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2 + 1)} \right\} &= \int_{0}^{t} \sin \tau d\tau = 1 - \cos t \\
\mathcal{L}^{-1}\left\{ \frac{1}{s^2(s^2 + 1)} \right\} &= \int_{0}^{t} (1 - \cos \tau) d\tau = t - \sin t
\end{aligned}
\)
Method 2: Use Partial Fraction Decomposition.
\(
\begin{aligned}
F(s) &= \frac{1}{s^2} - \frac{1}{s^2 + 1}
\end{aligned}
\)
Conclusion: \[ f(t) = t - \sin t \]