Question

Find the laplace transform of the following function f(t)=6sin2t-5cos2t

Answer 1

Step-by-step explanation:

Given Find the laplace transform of the following function f(t)=6 sin 2t-5 cos 2t

• L(6 sin 2t – 5 cos 2t) = 0 to infinity ʃ e^-st (e sin 2t – 5 cos 2t)dt
• Now I  = 0 to infinity ʃ sin 2t dt = [- 1/s e^-st sin 2t] 0 to infinity + 0 to infinity ʃ 2/s cos 2t dt
•                              = 0 + [-2/s^2 e^-st cos 2t]0 to infinity + 0 to infinity ʃ - 4 / s^2 sin 2t dt
•                             = 2/s^2 – 4/s^2 I
• Therefore s^2 + 4 / s^2 = 2/s^2  
• So I = 2/s^2 + 4

Similarly 0 to infinity cos2t dt = s/s^2 + 4 and so L(6sin2t – 5cos2t) = 12 – 5s /s^2 + 4 for s > 0

Answer 2

The Laplace transform of f(t) = 6 sin2t - 5 cos2t is (2 - 5s)/(s² + 4)

Step-by-step explanation:

$\mathcal{L}\{6 \sin 2 t-5 \cos 2 t\}=\int_{0}^{\infty} e^{-s t}\{6 \sin 2 t-5 \cos 2 t\} d t$

We need to integrate separately,

Let $\int_{0}^{\infty} \sin 2 t d t$ be I

$\int_{0}^{\infty} \sin 2 t d t=\left[-\frac{1}{s} e^{-s t} \sin 2 t\right]_{0}^{\infty}+\int_{0}^{\infty} \frac{2}{s} \cos 2 t d t$

$\int_{0}^{\infty} \sin 2 t d t=0+\left[-\frac{2}{s^{2}} e^{-s t} \cos 2 t\right]_{0}^{\infty}+\int_{0}^{\infty} \frac{-4}{s^{2}} \sin 2 t d t$

$\int_{0}^{\infty} \sin 2 t d t=\frac{2}{s^{2}}-\frac{4}{s^{2}} I$

$\frac{s^{2}+4}{s^{2}} I=\frac{2}{s^{2}}$

Now,

$I=\frac{2}{s^{2}+4}$

Similarly, on integrating another part of the given function, we get,

$\int_{0}^{\infty} \cos 2 t d t=\frac{s}{s^{2}+4}$

Now, the Laplace transform is:

$\therefore \mathcal{L}\{6 \sin 2 t-5 \cos 2 t\}=\frac{2-5 s}{s^{2}+4}$