# 8.2E: The Inverse Laplace Transform (Exercises)

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## Q8.2.1

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

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

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

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

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

4. Find the inverse Laplace transform.

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

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

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

6. Find the inverse Laplace transform.

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

7. Find the inverse Laplace transform.

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

8. Find the inverse Laplace transform.

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

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.

This page titled 8.2E: The Inverse Laplace Transform (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.