Question

find the laplace tranform of the following
1-(t+5)^2cos10t
2-te^-10t sin^2 t

Answer 1

Answer on Question #42600 – Math – Differential Calculus

Find the laplace transform of the following $1-(t+5)^2\cos10t$, $2-\text{te}^-10t \sin^2 t$

Answer.

1. $f(t) = 1 - (t + 5)^2 \cos 10t$

$$F(s) = L_t\{f(t)\}(s) = \int_0^\infty [1 - (t + 5)^2 \cos 10t] e^{-st} dt =$$ $$\int_0^\infty [1 - t^2 \cos 10t - 10t \cos 10t - 25 \cos 10t] e^{-st} dt,$$

and using well known Laplace transforms:

$$L_t\{1\}(s) = \frac{1}{s}. \quad L_t\{\cos(at)\}(s) = \frac{s}{s^2 + a^2},$$ $$L_t\{t^n f(t)\}(s) = (-1)^n F^{(n)}(s), \quad \text{(where } F(s) = L_t\{f(t)\}(s)),$$ $$\text{so } L_t\{t \cos(at)\}(s) = \frac{s^2 - a^2}{(s^2 + a^2)^2}, \quad L_t\{t^2 \cos(at)\}(s) = \frac{2s(s^2 - 3a^2)}{(s^2 + a^2)^3},$$

finally, we have

$$F(s) = L_t\{f(t)\}(s) = \frac{1}{s} - \frac{25s}{s^2 + 100} - \frac{10(s^2 - 100)}{(s^2 + 100)^2} - \frac{2s(s^2 - 300)}{(s^2 + 100)^3}.$$

2. $f(t) = 2 - t e^{-10t} \sin 2t$

$$F(s) = L_t\{f(t)\}(s) = \int_0^\infty [2 - t e^{-10t} \sin 2t] e^{-st} dt$$

and using well known Laplace transforms:

$$L_t\{1\}(s) = \frac{1}{s}. \quad L_t\{e^{bt} \sin(at)\}(s) = \frac{a}{(s - b)^2 + a^2},$$ $$L_t\{t f(t)\}(s) = -F'(s), \quad \text{(where } F(s) = L_t\{f(t)\}(s)),$$ $$\text{so } L_t\{t e^{bt} \sin(at)\}(s) = -\frac{2a(s - b)}{((s - b)^2 + a^2)^2},$$ $$\text{finally, we have } F(s) = L_t\{f(t)\}(s) = \frac{2}{s} - \frac{4(s + 10)}{(s^2 + 20s + 104)^2}.$$

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