MATH 308 Lecture 21
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Theorem 6.3.1
If exists for and if is a positive constant, then
Exercise 4
Find the Laplace transform of
Rewrite
For second term, let , then
For third term, let , then
Then we have
Theorem 6.3.2
If exists for and if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is a constant, then
Conversely, if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(t) = \mathcal{L}^{-1} \left\{ F(s) \right\}} , then
Exercise 5
Find the inverse Laplace transform of the functions:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(s) = \frac{\mathrm{e}^{-2x}}{s^2-2s-3}}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(s) = \frac{\mathrm{e}^{-s}-\mathrm{e}^{-3s}+3}{s}}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H(s) = \frac{2 \mathrm{e}^{-3s}}{(s-1)^2 + 4}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathcal{L}^{-1} \left\{ G(s) \right\} &= \mathcal{L}^{-1} \left\{ \frac{\mathrm{e}^{-s}}{s} - \frac{\mathrm{e}^{-3s}}{s} + \frac{3}{s} \right\} \\ &= u_1(t) - u_3(t) + 3 \end{align}}
Exercise 7
Find solution of initial value problems Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y''+y = u_{3\pi}(t)} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(0) = 0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'(0) = 1}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathcal{L} \left\{ y'' \right\} + \mathcal{L} \left\{ y \right\} &= \mathcal{L} \left\{ 1 - u_{3\pi}(t) \right\} \\ s^2 \mathcal{L} \left\{ y \right\} - y'(0) - s \, y(0) + \mathcal{L} \left\{ y \right\} &= \frac{1-\mathrm{e}^{-3\pi\,s}}{s} \\ (s^2 + 1) \mathcal{L} \left\{ y \right\} - 1 &= \frac{1-\mathrm{e}^{-3\pi\,s}}{s} \end{align}}