8.2E: The Inverse Laplace Transform (Exercises)

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Jul 20, 2020
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- 8.2: The Inverse Laplace Transform
- 8.3: Solution of Initial Value Problems

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Q8.2.1

1. Use the table of Laplace transforms to find the inverse Laplace transform.

$\dfrac{3}{(s-7)^4}$
$\dfrac{2s-4}{s^2-4s+13}$
$\dfrac{1}{s^2+4s+20}$
$\dfrac{2}{s^2+9}$
$\dfrac{s^2-1}{(s^2+1)^2}$
$\dfrac{1}{(s-2)^2-4}$
$\dfrac{12s-24}{(s^2-4s+85)^2}$
$\dfrac{2}{(s-3)^2-9}$
$\dfrac{s^2-4s+3}{(s^2-4s+5)^2}$

2. Use Theorem 8.2.1 and the table of Laplace transforms to find the inverse Laplace transform.

$\dfrac{2s+3}{(s-7)^4}$
$\dfrac{s^2-1}{(s-2)^6}$
$\dfrac{s+5}{ s^2+6s+18}$
$\dfrac{2s+1}{ s^2+9}$
$\dfrac{s}{ s^2+2s+1}$
$\dfrac{s+1}{ s^2-9}$
$\dfrac{s^3+2s^2-s-3}{(s+1)^4}$
$\dfrac{2s+3}{(s-1)^2+4}$
$\dfrac{1}{ s}-\dfrac{s}{ s^2+1}$
$\dfrac{3s+4}{ s^2-1}$
$\dfrac{3}{ s-1}+\dfrac{4s+1}{ s^2+9}$
$\dfrac{3}{(s+2)^2}-\dfrac{2s+6}{ s^2+4}$

3. Use Heaviside’s method to find the inverse Laplace transform.

$\dfrac{3-(s+1)(s-2)}{(s+1)(s+2)(s-2)}$
$\dfrac{7+(s+4)(18-3s)}{(s-3)(s-1)(s+4)}$
$\dfrac{2+(s-2)(3-2s)}{(s-2)(s+2)(s-3)}$
$\dfrac{3-(s-1)(s+1)}{(s+4)(s-2)(s-1)}$
$\dfrac{3+(s-2)(10-2s-s^2)}{(s-2)(s+2)(s-1)(s+3)}$
$\dfrac{3+(s-3)(2s^2+s-21)}{(s-3)(s-1)(s+4)(s-2)}$

4. Find the inverse Laplace transform.

$\dfrac{2+3s}{(s^2+1)(s+2)(s+1)}$
$\dfrac{3s^2+2s+1}{(s^2+1)(s^2+2s+2)}$
$\dfrac{3s+2}{(s-2)(s^2+2s+5)}$
$\dfrac{3s^2+2s+1}{(s-1)^2(s+2)(s+3)}$
$\dfrac{2s^2+s+3}{(s-1)^2(s+2)^2}$
$\dfrac{3s+2}{(s^2+1)(s-1)^2}$

5. Use the method of Example 8.2.9 to find the inverse Laplace transform.

$\dfrac{3s+2}{(s^2+4)(s^2+9)}$
$\dfrac{-4s+1}{(s^2+1)(s^2+16)}$
$\dfrac{5s+3}{(s^2+1)(s^2+4)}$
$\dfrac{-s+1}{(4s^2+1)(s^2+1)}$
$\dfrac{17s-34}{(s^2+16)(16s^2+1)}$
$\dfrac{2s-1}{(4s^2+1)(9s^2+1)}$

6. Find the inverse Laplace transform.

$\dfrac{17 s-15}{(s^2-2s+5)(s^2+2s+10)}$
$\dfrac{8s+56}{(s^2-6s+13)(s^2+2s+5)}$
$\dfrac{s+9}{(s^2+4s+5)(s^2-4s+13)}$
$\dfrac{3s-2}{(s^2-4s+5)(s^2-6s+13)}$
$\dfrac{3s-1}{(s^2-2s+2)(s^2+2s+5)}$
$\dfrac{20s+40}{(4s^2-4s+5)(4s^2+4s+5)}$

7. Find the inverse Laplace transform.

$\dfrac{1}{ s(s^2+1)}$
$\dfrac{1}{(s-1)(s^2-2s+17)}$
$\dfrac{3s+2}{(s-2)(s^2+2s+10)}$
$\dfrac{34-17s}{(2s-1)(s^2-2s+5)}$
$\dfrac{s+2}{(s-3)(s^2+2s+5)}$
$\dfrac{2s-2}{(s-2)(s^2+2s+10)}$

8. Find the inverse Laplace transform.

$\dfrac{2s+1}{(s^2+1)(s-1)(s-3)}$
$\dfrac{s+2}{(s^2+2s+2)(s^2-1)}$
$\dfrac{2s-1}{(s^2-2s+2)(s+1)(s-2)}$
$\dfrac{s-6}{(s^2-1)(s^2+4)}$
$\dfrac{2s-3}{ s(s-2)(s^2-2s+5)}$
$\dfrac{5s-15}{(s^2-4s+13)(s-2)(s-1)}$

9. Given that $f(t)\leftrightarrow F(s)$ , find the inverse Laplace transform of $F(as-b)$ , where $a>0$ .

1. If $s_1$ , $s_2$ , …, $s_n$ are distinct and $P$ is a polynomial of degree less than $n$ , then ${P(s)}{(s-s_1)(s-s_2)\cdots(s-s_n)}= {A_1}{ s-s_1}+{A_2}{ s-s_2}+\cdots+{A_n}{ s-s_n}.\nonumber$ Multiply through by $s-s_i$ to show that $A_i$ can be obtained by ignoring the factor $s-s_i$ on the left and setting $s=s_i$ elsewhere.
2. Suppose $P$ and $Q_1$ are polynomials such that $\mbox{degree}(P)\le\mbox{degree}(Q_1)$ and $Q_1(s_1)\ne0$ . Show that the coefficient of $1/(s-s_1)$ in the partial fraction expansion of $F(s)={P(s)}{(s-s_1)Q_1(s)}\nonumber$ is $P(s_1)/Q_1(s_1)$ .
3. Explain how the results of (a) and (b) are related.

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Q8.2.1

Q8.2.1

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