Question

Find the Laplace transform te.​

Answer 1

Answer:

According to the first shifting theorem,

$\text{If Laplace transform of f(t) = F(s), then the Laplace Transform of:}\\\\\\\boxed{ \bf{(e^{at}.f(t)) = F(s-a)}}$

According to the question,

$\implies f(t) = t.e^{-t}\\\\\\\text{Comparing with the theorem, we get:}\\\\\implies e^{at} = e^{-t}\\\\\implies \bf{a=-1}$

Also f(t) = t.

Calculating the Laplace transform of (t) we get:

$\implies L(t) = \dfrac{1}{s^2}\\\\\\\implies F(s-a) = \dfrac{1}{(s-a)^2}\\\\\\\implies F(s) = \dfrac{1}{(s-(-1))^2}\\\\\\\implies \boxed{ \bf{ L(t.e^{-t}) = \dfrac{1}{(s+1)^2}}}$

Hence this is the required answer.

Answer 2

According to the first shifting theorem,

If Laplace transform of f(t) = F(s), then the Laplace Transform of:

$\tiny\sf \: ({e}^{at}.f(t)) = F(s - a)$

According to the question,

$\tiny\sf \:➻ \: f(t) = {t.e}^{ - t}$

Comparing with theorem, we get :

$\tiny\sf \:➻ \: {e}^{at} = {e}^{ - t}$

$\tiny\sf \:➻ \: a \: = - 1$

Also f(t) = t.

Calculating the Laplace transform of (t) we get :

$\tiny\sf \:➻ L(t) = \frac{1}{ {s}^{2} }$

$\tiny\sf \:➻ F(s - a) = \frac{1}{(s - a)²}$