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Lana71 [14]
3 years ago
11

The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

r integral L{f(t)}(s) = infinity 0 e-st.f(t)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of as a fixed constant)
1. Find L{t}. (hint: remember integration by parts)
A. 1
B. -1/s2
C. 0
D. 1/s2
E. -s2
F. None of these
2. Find L{1}.
a.1/s
b. 1
c. 0
d. -s
e. -1/s
f. none of these
Mathematics
1 answer:
lesya692 [45]3 years ago
7 0

(1) D

L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt

Integrate by parts, taking

u = t \implies \mathrm du=\mathrm dt

\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}

Then

L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}

L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}

(2) A

L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}

L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}

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