Multiplication by \( t^n \) & Division by \( t \)

Find the Laplace transforms of the following:

1. \( t \sin^2(t) \)

Solution: \[ L\{ t \sin^2(t) \} = L\left\{ t \cdot \frac{1 - \cos(2t)}{2} \right\} = \frac{1}{2} \left[ L\{ t \} - L\{ t \cos(2t) \} \right]. \] Using the multiplication by \( t \) property: \[ L\{ t \cos(2t) \} = -\frac{d}{ds} \left( \frac{s}{s^2 + 4} \right). \] Simplifying: \[ L\{ t \sin^2(t) \} = \frac{1}{2} \left[ \frac{1}{s^2} + \frac{d}{ds} \left( \frac{s}{s^2 + 4} \right) \right]. \] Final result: \[ L\{ t \sin^2(t) \} = \frac{2(3s^2 + 4)}{s^2 (s^2 + 4)^2}. \]

2. \( t e^{-2t} \sin(4t) \)

Solution: Using the first shifting theorem and multiplication by \( t \): \[ L\{ t e^{-2t} \sin(4t) \} = \left[ L\{ t \sin(4t) \} \right]_{s \to (s + 2)}. \] Simplifying: \[ L\{ t e^{-2t} \sin(4t) \} = \frac{8(s + 2)}{(s^2 + 4s + 20)^2}. \]

Division by \( t \):

If \( L\{ f(t) \} = F(s) \), then: \[ L\left\{ \frac{f(t)}{t} \right\} = \int_{s}^{\infty} F(s) \, ds, \] provided that the integral exists.

Find the Laplace transforms of the following

1. \( \frac{\sin(t)}{t} \)

Solution: \[ L\left\{ \frac{\sin(t)}{t} \right\} = \int_{s}^{\infty} L\{ \sin(t) \} \, ds = \int_{s}^{\infty} \frac{1}{s^2 + 1} \, ds. \] Evaluating the integral: \[ L\left\{ \frac{\sin(t)}{t} \right\} = \left[ \tan^{-1}(s) \right]_{s}^{\infty} = \frac{\pi}{2} - \tan^{-1}(s) = \cot^{-1}(s). \]

2. \( \frac{e^{at} - \cos(bt)}{t} \)

Solution: \[ L\left\{ \frac{e^{at} - \cos(bt)}{t} \right\} = \int_{s}^{\infty} L\{ e^{at} - \cos(bt) \} \, ds. \] Substituting the Laplace transforms: \[ = \int_{s}^{\infty} \left[ \frac{1}{s - a} - \frac{s}{s^2 + b^2} \right] \, ds. \] Evaluating the integral: \[ = \left[ \log(s - a) - \frac{1}{2} \log(s^2 + b^2) \right]_{s}^{\infty}. \] Simplifying: \[ = \log\left( \frac{\sqrt{s^2 + b^2}}{s - a} \right). \]

Model Problems

Find the Laplace transform of the following functions:

1. \[ \sin(2t) - 2t \cos(2t) \]
2. \[ t^2 e^{t} \sin(t) \]
3. \[ t \sinh(at) \]
4. \[ t \cos(at) \]
5. \[ t^2 \sin(at) \]
6. \[ t e^{-t} \sin(3t) \]
7. \[ t^2 e^{-3t} \sin(2t) \]
8. \[ t \cos(3t) \cos(2t) \]
9. \[ (t^2 - 3t + 2) \sin(3t) \]
10. \[ \frac{1 - \cos(3t)}{t} \]
11. \[ \frac{\cos(4t) \sin(2t)}{t} \]
12. \[ \frac{1 - e^{t}}{t} \]
13. \[ \frac{e^{-at} - e^{-bt}}{t} \]
14. \[ \frac{e^{-3t} \sin(2t)}{t} \]
15. \[ \frac{\cos(at) - \cos(bt)}{t} \]
16. \[ \frac{\sin(t) \sin(5t)}{t} \]
17. \[ \frac{1 - \cos(t)}{t^2} \]

Answers

1. \[ \frac{16}{(s^2 + 4)^2} \]
2. \[ \frac{2(3s^2 - 6s + 2)}{(s^2 - 2s + 2)^3} \]
3. \[ \frac{2as}{(s^2 - a^2)^2} \]
4. \[ \frac{s^2 - a^2}{(s^2 + a^2)^2} \]
5. \[ \frac{2a(3s^2 - a^2)}{(s^2 + a^2)^3} \]
6. \[ \frac{6(s + 1)}{(s^2 + 2s + 10)^2} \]
7. \[ \frac{4(3s^2 + 18s + 23)}{(s^2 + 6s + 13)^3} \]
8. \[ \frac{1}{2} \left[ \frac{s^2 - 25}{(s^2 + 25)^2} + \frac{s^2 - 1}{(s^2 + 1)^2} \right] \]
9. \[ \frac{6s^4 - 18s^3 + 126s^2 - 162s + 432}{(s^2 + 9)^3} \]
10. \[ \log \left( \frac{\sqrt{s^2 + 9}}{s} \right) \]
11. \[ \frac{1}{2} \left[ \tan^{-1}\left( \frac{s}{2} \right) - \tan^{-1}\left( \frac{s}{6} \right) \right] \]
12. \[ \log \left( \frac{s - 1}{s} \right) \]
13. \[ \log \left( \frac{s + b}{s + a} \right) \]
14. \[ \cot^{-1}\left( \frac{s + 3}{2} \right) \]
15. \[ \frac{1}{2} \log \left( \frac{s^2 + b^2}{s^2 + a^2} \right) \]
16. \[ \frac{1}{4} \log \left( \frac{s^2 + 36}{s^2 + 16} \right) \]
17. \[ \cot^{-1}(s) - \frac{s}{2} \log \left( 1 + \frac{1}{s^2} \right) \]