Section 8.1 Introduction to the Laplace Transforms

Problem 1.
Find the Laplace transforms of the following functions by evaluating the integral
[tex]F(s) = {∫}^{ \infty } _{0} {e}^{ - st} f(t)dt[/tex]
[tex](a)t[/tex]
[tex](b) {te}^{ - t} [/tex]
[tex](c) {sinh} \: bt[/tex]
[tex](d) {e}^{2t} - {3e}^{t} [/tex]
[tex](e) {t}^{2} [/tex]

​

For the integrals in (a), (b), and (e), you'll end up integrating by parts.

(a)

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt[/tex]

Let

[tex]u = t \implies du = dt[/tex]

[tex]dv = e^{-st} \, dt \implies v = -\dfrac1s e^{-st}[/tex]

Then

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = uv\bigg|_{t=0}^{t\to\infty} - \int_0^\infty v\, du[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = \left(-\frac1s te^{-st}\right)\bigg|_0^\infty + \frac1s \int_0^\infty e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1s \left(\lim_{t\to\infty} te^{-st} - 0\right) + \frac1s \int_0^\infty e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = \frac1s \int_0^\infty e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1{s^2} e^{-st} \bigg|_0^\infty e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1{s^2} \left(\lim_{t\to\infty}e^{-st} - 1\right) = \boxed{\frac1{s^2}}[/tex]

(b)

[tex]\displaystyle \int_0^\infty t e^{-t} e^{-st} \, dt = \int_0^\infty t e^{-(s+1)t} \, dt[/tex]

Let

[tex]u = t \implies du = dt[/tex]

[tex]dv = e^{-(s+1)t} \, dt \implies v = -\dfrac1{s+1} e^{-(s+1)}t[/tex]

Then

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\dfrac1{s+1} te^{-(s+1)t} \bigg|_0^\infty + \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\dfrac1{s+1} \left(\lim_{t\to\infty}te^{-(s+1)t} - 0\right) + \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\frac1{(s+1)^2} e^{-(s+1)t} \bigg|_0^\infty[/tex]

[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\frac1{(s+1)^2} \left(\lim_{t\to\infty}e^{-(s+1)t} - 1\right) = \boxed{\frac1{(s+1)^2}}[/tex]

(e)

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt[/tex]

Let

[tex]u = t^2 \implies du = 2t \, dt[/tex]

[tex]dv = e^{-st} \, dt \implies v = -\dfrac1s e^{-st}[/tex]

Then

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = -\frac1s t^2 e^{-st} \bigg|_0^\infty + \frac2s \int_0^\infty t e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = -\frac1s \left(\lim_{t\to\infty} t^2 e^{-st} - 0\right) + \frac2s \int_0^\infty t e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = \frac2s \int_0^\infty t e^{-st} \, dt[/tex]

The remaining integral is the transform we found in (a), so

[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = \frac2s \times \frac1{s^2} = \boxed{\frac2{s^3}}[/tex]

Computing the integrals in (c) and (d) is much more immediate.

(c)

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \int_0^\infty \frac{e^{bt}-e^{-bt}}2 \times e^{-st} \, dt[/tex]

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \int_0^\infty \left(e^{(b-s)t} - e^{(b+s)t}\right) \, dt[/tex]

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left(\frac1{b-s} e^{(b-s)t} - \frac1{b+s} e^{(b+s)t}\right) \bigg|_0^\infty[/tex]

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left[\lim_{t\to\infty}\left(\frac1{b-s} e^{(b-s)t} - \frac1{b+s} e^{(b+s)t}\right) - \left(\frac1{b-s} - \frac1{b+s}\right)\right][/tex]

[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left(\frac1{b+s} - \frac1{b-s}\right) = \boxed{\frac{s}{s^2-b^2}}[/tex]

(d)

[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \int_0^\infty \left(e^{(2-s)t} - 3e^{(1-s)t}\right) \, dt[/tex]

[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \left( \frac1{2-s} e^{(2-s)t} - \frac3{1-s} e^{(1-s)t} \right) \bigg|_0^\infty[/tex]

[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \lim_{t\to\infty} \left( \frac1{2-s} e^{(2-s)t} - \frac3{1-s} e^{(1-s)t} \right) - \left( \frac1{2-s} - \frac3{1-s} \right)[/tex]

[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \frac3{1-s} - \frac1{2-s} = \boxed{-\frac{2s-5}{s^2-3s+2}}[/tex]

Simplify the expression.

5(10 + g) + 14g + 4
A.
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B.
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C.
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D.
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