Inverse Laplace Transforms

Contents

L.T. of Elementary Functions

Multiplication by tn & Division by t

L.T. of Integrals and Special Functions

Evaluation of Integrals using L.T

Inverse Laplace Transforms

Convolution Theorem

Application to Differential Equations

1M Questions

Definition:
If \( L\{ f(t) \} = F(s) \), then \( f(t) \) is known as the inverse Laplace transform of \( F(s) \) and is denoted by: \[ f(t) = L^{-1}\{ F(s) \}. \]

Table of Inverse Laplace Transforms

To save effort in determining the inverse Laplace transform of a given function \( F(s) \), a look-up table is provided below, listing some common functions and their corresponding inverse Laplace transforms.

S.No. \( F(s) \) \( L^{-1}\{ F(s) \} = f(t) \)
1 \( \frac{1}{s} \) \( 1 \)
2 \( \frac{1}{s^2} \) \( t \)
3 \( \frac{1}{s^n}, \, n \in \mathbb{Z}^+ \) \( \frac{t^{n-1}}{(n-1)!} \)
4 \( \frac{1}{s - a} \) \( e^{at} \)
5 \( \frac{1}{s + a} \) \( e^{-at} \)
6 \( \frac{s}{s^2 + a^2} \) \( \cos(at) \)
7 \( \frac{1}{s^2 + a^2} \) \( \frac{1}{a} \sin(at) \)
8 \( \frac{s}{s^2 - a^2} \) \( \cosh(at) \)
9 \( \frac{1}{s^2 - a^2} \) \( \frac{1}{a} \sinh(at) \)
10 \( \frac{s}{(s^2 - a^2)^2} \) \( \frac{t}{2a} \sin(at) \)
11 \( \frac{1}{(s^2 + a^2)^2} \) \( \frac{1}{2a^3} (\sin(at) - at \cos(at)) \)

General Properties of Inverse Laplace Transforms

  1. Linearity Property: If \( L^{-1}\{ F(s) \} = f(t) \) and \( L^{-1}\{ G(s) \} = g(t) \), then: \[ L^{-1}\{ c_1 F(s) + c_2 G(s) \} = c_1 f(t) + c_2 g(t), \] where \( c_1 \) and \( c_2 \) are constants.
  2. First Shifting Theorem: If \( L^{-1}\{ F(s) \} = f(t) \), then: \[ L^{-1}\{ F(s - a) \} = e^{at} f(t). \]
  3. Change of Scale Property: If \( L^{-1}\{ F(s) \} = f(t) \), then: \[ L^{-1}\{ F(as) \} = \frac{1}{a} f\left( \frac{t}{a} \right). \]
  4. Second Shifting Theorem: If \( L^{-1}\{ F(s) \} = f(t) \), then: \[ L^{-1}\{ e^{-as} F(s) \} = \begin{cases} f(t - a), & t > a, \\ 0, & t < a. \end{cases} \]
  5. Derivative Property: \[ L^{-1}\left\{ \frac{d}{ds} [F(s)] \right\} = -t L^{-1}\{ F(s) \}. \]

Method of Partial Fractions to Find Inverse Laplace Transform

If \( F(s) = \frac{\phi(s)}{\psi(s)} \), where \( \phi \) and \( \psi \) are polynomials in \( s \), then \( f(t) \) can be obtained by resolving \( F(s) \) into partial fractions and manipulating term by term.

Steps:

  1. Factorize the denominator \( \psi(s) \) into linear or quadratic factors.
  2. Express \( F(s) \) as a sum of simpler fractions (partial fractions).
  3. Use the linearity property and the inverse Laplace transform table to find \( f(t) \).

1. Evaluate \[ L^{-1}\left\{ \frac{3(s^2 - 2)^2}{2s^5} \right\}. \]

Solution: \[ L^{-1}\left\{ \frac{3(s^2 - 2)^2}{2s^5} \right\} = \frac{3}{2} L^{-1}\left\{ \frac{(s^2 - 2)^2}{s^5} \right\}. \] Expand the numerator: \[ \frac{(s^2 - 2)^2}{s^5} = \frac{s^4 - 4s^2 + 4}{s^5} = \frac{1}{s} - \frac{4}{s^3} + \frac{4}{s^5}. \] Apply the inverse Laplace transform term by term: \[ L^{-1}\left\{ \frac{1}{s} \right\} = 1, \quad L^{-1}\left\{ \frac{1}{s^3} \right\} = \frac{t^2}{2!}, \quad L^{-1}\left\{ \frac{1}{s^5} \right\} = \frac{t^4}{4!}. \] Combine the results: \[ L^{-1}\left\{ \frac{3(s^2 - 2)^2}{2s^5} \right\} = \frac{3}{2} \left[ 1 - 4 \cdot \frac{t^2}{2!} + 4 \cdot \frac{t^4}{4!} \right]. \] Simplify: \[ = \frac{3}{2} \left[ 1 - 2t^2 + \frac{t^4}{6} \right] = \frac{1}{4} \left[ t^4 - 12t^2 + 6 \right]. \]

2. Evaluate \[ L^{-1}\left\{ \frac{1}{s^2 - 5s + 6} \right\}. \]

Solution: Factorize the denominator: \[ \frac{1}{s^2 - 5s + 6} = \frac{1}{(s - 2)(s - 3)}. \] Use partial fractions: \[ \frac{1}{(s - 2)(s - 3)} = \frac{1}{s - 3} - \frac{1}{s - 2}. \] Apply the inverse Laplace transform: \[ L^{-1}\left\{ \frac{1}{s - 3} \right\} = e^{3t}, \quad L^{-1}\left\{ \frac{1}{s - 2} \right\} = e^{2t}. \] Combine the results: \[ L^{-1}\left\{ \frac{1}{s^2 - 5s + 6} \right\} = e^{3t} - e^{2t}. \]

3. Evaluate \[ L^{-1}\left\{ \frac{s}{(s^2 + 1)(s^2 + 9)(s^2 + 25)} \right\}. \]

Solution: Consider: \[ \frac{1}{(s^2 + 1)(s^2 + 9)(s^2 + 25)} = \frac{1}{(p + 1)(p + 9)(p + 25)}, \quad \text{where } p = s^2. \] Using partial fractions: \[ \frac{1}{(p + 1)(p + 9)(p + 25)} = \frac{A}{p + 1} + \frac{B}{p + 9} + \frac{C}{p + 25}. \] Solve for \( A \), \( B \), and \( C \): \[ 1 = A(p + 9)(p + 25) + B(p + 1)(p + 25) + C(p + 1)(p + 9). \] Substitute \( p = -1 \): \[ 1 = A(8)(24) \Rightarrow A = \frac{1}{192}. \] Substitute \( p = -9 \): \[ 1 = B(-8)(16) \Rightarrow B = -\frac{1}{128}. \] Substitute \( p = -25 \): \[ 1 = C(-24)(-16) \Rightarrow C = \frac{1}{384}. \] Substituting \( A \), \( B \), and \( C \): \[ \frac{1}{(s^2 + 1)(s^2 + 9)(s^2 + 25)} = \frac{1}{192(s^2 + 1)} - \frac{1}{128(s^2 + 9)} + \frac{1}{384(s^2 + 25)}. \] Multiply by \( s \) and apply the inverse Laplace transform: \[ L^{-1}\left\{ \frac{s}{(s^2 + 1)(s^2 + 9)(s^2 + 25)} \right\} = \frac{1}{192} L^{-1}\left\{ \frac{s}{s^2 + 1} \right\} - \frac{1}{128} L^{-1}\left\{ \frac{s}{s^2 + 9} \right\} + \frac{1}{384} L^{-1}\left\{ \frac{s}{s^2 + 25} \right\}. \] Using the inverse Laplace transform table: \[ L^{-1}\left\{ \frac{s}{s^2 + 1} \right\} = \cos t, \quad L^{-1}\left\{ \frac{s}{s^2 + 9} \right\} = \cos 3t, \quad L^{-1}\left\{ \frac{s}{s^2 + 25} \right\} = \cos 5t. \] Therefore: \[ L^{-1}\left\{ \frac{s}{(s^2 + 1)(s^2 + 9)(s^2 + 25)} \right\} = \frac{1}{192} \cos t - \frac{1}{128} \cos 3t + \frac{1}{384} \cos 5t. \]

4. Evaluate \[ L^{-1}\left\{ \frac{s}{(s - 3)(s^2 + 4)} \right\}. \]

Solution: Use partial fractions to decompose the function: \[ \frac{s}{(s - 3)(s^2 + 4)} = \frac{A}{s - 3} + \frac{Bs + C}{s^2 + 4}. \] Solve for \( A \), \( B \), and \( C \): \[ s = A(s^2 + 4) + (Bs + C)(s - 3). \] Substitute \( s = 3 \): \[ 3 = A(9 + 4) \Rightarrow A = \frac{3}{13}. \] Substitute \( s = 0 \): \[ 0 = A(4) + C(-3) \Rightarrow 0 = \frac{12}{13} - 3C \Rightarrow C = \frac{4}{13}. \] Compare the coefficient of \( s^2 \): \[ 0 = A + B \Rightarrow B = -A = -\frac{3}{13}. \] Rewrite the partial fractions: \[ \frac{s}{(s - 3)(s^2 + 4)} = \frac{3}{13(s - 3)} - \frac{3s}{13(s^2 + 4)} + \frac{4}{13(s^2 + 4)}. \] Apply the inverse Laplace transform: \[ L^{-1}\left\{ \frac{s}{(s - 3)(s^2 + 4)} \right\} = \frac{3}{13} L^{-1}\left\{ \frac{1}{s - 3} \right\} - \frac{3}{13} L^{-1}\left\{ \frac{s}{s^2 + 4} \right\} + \frac{4}{13} L^{-1}\left\{ \frac{1}{s^2 + 4} \right\}. \] Using the inverse Laplace transform table: \[ L^{-1}\left\{ \frac{1}{s - 3} \right\} = e^{3t}, \quad L^{-1}\left\{ \frac{s}{s^2 + 4} \right\} = \cos 2t, \quad L^{-1}\left\{ \frac{1}{s^2 + 4} \right\} = \frac{1}{2} \sin 2t. \] Therefore: \[ L^{-1}\left\{ \frac{s}{(s - 3)(s^2 + 4)} \right\} = \frac{3}{13} e^{3t} - \frac{3}{13} \cos 2t + \frac{4}{13} \cdot \frac{1}{2} \sin 2t. \] Simplify: \[ = \frac{3}{13} e^{3t} - \frac{3}{13} \cos 2t + \frac{2}{13} \sin 2t = \frac{1}{13} \left( 3e^{3t} - 3\cos 2t + 4\sin 2t \right). \]

5. Evaluate \[ L^{-1}\left\{ \frac{s + 2}{s^2 - 2s + 5} \right\}. \]

Solution: Rewrite the denominator: \[ s^2 - 2s + 5 = (s - 1)^2 + 4. \] Rewrite the numerator: \[ s + 2 = (s - 1) + 3. \] Therefore: \[ L^{-1}\left\{ \frac{s + 2}{s^2 - 2s + 5} \right\} = L^{-1}\left\{ \frac{(s - 1) + 3}{(s - 1)^2 + 4} \right\}. \] Using the first shifting theorem: \[ L^{-1}\left\{ \frac{(s - 1) + 3}{(s - 1)^2 + 4} \right\} = e^{t} L^{-1}\left\{ \frac{s + 3}{s^2 + 4} \right\}. \] Apply the inverse Laplace transform: \[ L^{-1}\left\{ \frac{s}{s^2 + 4} \right\} = \cos 2t, \quad L^{-1}\left\{ \frac{3}{s^2 + 4} \right\} = \frac{3}{2} \sin 2t. \] Combine the results: \[ L^{-1}\left\{ \frac{s + 2}{s^2 - 2s + 5} \right\} = e^{t} \left( \cos 2t + \frac{3}{2} \sin 2t \right). \]

6. Evaluate \[ L^{-1}\left\{ \frac{1 + e^{-\pi s}}{s^2 + 1} \right\}. \]

Solution: Split the expression: \[ L^{-1}\left\{ \frac{1 + e^{-\pi s}}{s^2 + 1} \right\} = L^{-1}\left\{ \frac{1}{s^2 + 1} \right\} + L^{-1}\left\{ \frac{e^{-\pi s}}{s^2 + 1} \right\}. \] Let \( F(s) = \frac{1}{s^2 + 1} \), then: \[ L^{-1}\left\{ \frac{1}{s^2 + 1} \right\} = \sin t = f(t). \] Using the second shifting theorem: \[ L^{-1}\left\{ e^{-\pi s} F(s) \right\} = \begin{cases} f(t - \pi), & t > \pi, \\ 0, & t < \pi. \end{cases} \] Therefore: \[ L^{-1}\left\{ \frac{e^{-\pi s}}{s^2 + 1} \right\} = \sin(t - \pi) \cdot u(t - \pi) = -\sin t \cdot u(t - \pi). \] Combine the results: \[ L^{-1}\left\{ \frac{1 + e^{-\pi s}}{s^2 + 1} \right\} = \sin t - \sin t \cdot u(t - \pi). \]

7. Evaluate \[ L^{-1}\left\{ \log \left( \frac{1 + s}{s} \right) \right\}. \]

Solution: Using the derivative property of inverse Laplace transforms: \[ L^{-1}\left\{ \frac{d}{ds} [F(s)] \right\} = -t L^{-1}\{ F(s) \}. \] Let \( F(s) = \log \left( \frac{1 + s}{s} \right) \). Then: \[ \frac{d}{ds} [F(s)] = \frac{d}{ds} \left[ \log(1 + s) - \log s \right] = \frac{1}{1 + s} - \frac{1}{s}. \] Therefore: \[ L^{-1}\left\{ \frac{1}{1 + s} - \frac{1}{s} \right\} = e^{-t} - 1. \] Using the derivative property: \[ -t L^{-1}\{ F(s) \} = e^{-t} - 1. \] Solve for \( L^{-1}\{ F(s) \} \): \[ L^{-1}\left\{ \log \left( \frac{1 + s}{s} \right) \right\} = \frac{1 - e^{-t}}{t}. \]

8. Evaluate \[ L^{-1}\left\{ \cot^{-1} s \right\}. \]

Solution: Using the derivative property of inverse Laplace transforms: \[ L^{-1}\left\{ \frac{d}{ds} [F(s)] \right\} = -t L^{-1}\{ F(s) \}. \] Let \( F(s) = \cot^{-1} s \). Then: \[ \frac{d}{ds} [F(s)] = \frac{d}{ds} \left[ \cot^{-1} s \right] = -\frac{1}{1 + s^2}. \] Therefore: \[ L^{-1}\left\{ -\frac{1}{1 + s^2} \right\} = -\sin t. \] Using the derivative property: \[ -t L^{-1}\{ F(s) \} = -\sin t. \] Solve for \( L^{-1}\{ F(s) \} \): \[ L^{-1}\left\{ \cot^{-1} s \right\} = \frac{\sin t}{t}. \]

Model Problems

Find the inverse Laplace transforms of the following functions:

  1. \[ L^{-1}\left\{ \frac{2s - 5}{4s^2 + 25} + \frac{4s - 18}{9 - s^2} \right\} \]
  2. \[ L^{-1}\left\{ \frac{1}{s^2 - 5s + 6} \right\} \]
  3. \[ L^{-1}\left\{ \frac{s}{(2s - 1)(2s - 1)} \right\} \]
  4. \[ L^{-1}\left\{ \frac{1 - 7s}{(s - 3)(s - 1)(s + 2)} \right\} \]
  5. \[ L^{-1}\left\{ \frac{2s - 3}{s^2 + 4s + 13} \right\} \]
  6. \[ L^{-1}\left\{ \frac{s + 2}{(s^2 + 4s + 8)^2} \right\} \]
  7. \[ L^{-1}\left\{ \frac{s^2 + 2s + 3}{(s^2 + 2s + 2)(s^2 + 2s + 5)} \right\} \]
  8. \[ L^{-1}\left\{ \frac{s}{s^4 + 4a^4} \right\} \]
  9. \[ L^{-1}\left\{ \frac{e^{-2s}}{s^2 + 4s + 5} \right\} \]
  10. \[ L^{-1}\left\{ \frac{s e^{-3s}}{s^2 + 8s + 16} \right\} \]
  11. \[ L^{-1}\left\{ \log \left( \frac{s + a}{s + b} \right) \right\} \]
  12. \[ L^{-1}\left\{ \log \left( \frac{s^2 + 1}{(s - 1)^2} \right) \right\} \]
  13. \[ L^{-1}\left\{ s \log \left( \frac{s - 1}{s + 1} \right) \right\} \]
  14. \[ L^{-1}\left\{ \tan^{-1} \left( \frac{2}{s} \right) \right\} \]
  15. \[ L^{-1}\left\{ \frac{s^2 - 3s + 3}{s^3} \right\} \]
  16. \[ L^{-1}\left\{ \frac{4s + 15}{25 + 16s^2} \right\} \]
  17. \[ L^{-1}\left\{ \frac{4}{(s + 1)(s + 2)} \right\} \]
  18. \[ L^{-1}\left\{ \frac{2s^2 - 6s + 5}{s^3 - 6s^2 + 11s - 6} \right\} \]
  19. \[ L^{-1}\left\{ \frac{4s + 5}{(s - 1)^2 (s + 2)} \right\} \]
  20. \[ L^{-1}\left\{ \frac{5s + 3}{(s - 1)(s^2 + 2s + 5)} \right\} \]
  21. \[ L^{-1}\left\{ \frac{1}{s^3 (s^2 + 1)} \right\} \]
  22. \[ L^{-1}\left\{ \frac{1}{s^2 (s^2 + 1)(s^2 - 1)} \right\} \]
  23. \[ L^{-1}\left\{ \frac{s + 3}{s^2 - 10s + 29} \right\} \]
  24. \[ L^{-1}\left\{ \frac{s + 3}{(s^2 + 6s + 13)^2} \right\} \]
  25. \[ L^{-1}\left\{ \frac{3s + 1}{(s + 1)^4} \right\} \]
  26. \[ L^{-1}\left\{ \frac{3s + 1}{(s - 1)(s^2 + 1)} \right\} \]
  27. \[ L^{-1}\left\{ \cot^{-1}(s + 1) \right\} \]
  28. \[ L^{-1}\left\{ \tan^{-1} \left( \frac{2}{s^2} \right) \right\} \]
  29. \[ L^{-1}\left\{ \log \left( \frac{s + 1}{(s + 2)(s + 3)} \right) \right\} \]
  30. \[ L^{-1}\left\{ \log \left( \frac{s^2 + 1}{s(s + 1)} \right) \right\} \]
  31. \[ L^{-1}\left\{ \frac{1}{2} \log \left( \frac{s^2 + b^2}{s^2 + a^2} \right) \right\} \]
  32. \[ L^{-1}\left\{ \log \left( 1 + \frac{1}{s^2} \right) \right\} \]
  33. \[ L^{-1}\left\{ \frac{e^{-3s}}{(s - 4)^2} \right\} \]
  34. \[ L^{-1}\left\{ \frac{s e^{-s\pi / 2}}{(s - 4)^2} \right\} \]

Back to Top