2e^(3t) sin4t find laplace transformation?


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 128460 by BHOOPENDRA last updated on 07/Jan/21

$2 e^{3 t} s i n 4 t$ $f i n d l a p l a c e t r a n s f o r m a t i o n ?$
	Answered by Dwaipayan Shikari last updated on 07/Jan/21

	$L (2 e^{3 t} s i n 4 t)$ $= \frac{1}{i} \int_{0}^{\infty} (e^{3 t} e^{4 i t} - e^{- 4 i t} e^{3 t}) e^{- s t} d t$ $= \frac{1}{i} \int_{0}^{\infty} e^{- (s - 3 - 4 i) t} - e^{- (s + 4 i - 3) t} d t$ $= \frac{1}{i} (\frac{1}{(s - 3 - 4 i)} - \frac{1}{(s - 3 + 4 i)}) Γ (1)$ $= \frac{1}{i} (\frac{8 i}{(s^{2} - 6 s + 9 + 16)}) = \frac{8}{s^{2} - 6 s + 25}$
	Commented by BHOOPENDRA last updated on 07/Jan/21

	$t s i n h 2 t s i n 3 t$ $p l z h e l p m e o u t t h i s$ $n t h a n k s f o r t h i s s o l u t i o n$
	Answered by mnjuly1970 last updated on 07/Jan/21

	$s o l u t i o n :$ $L [f (t)] = F (s)$ $L [e^{a t} f (t)] = F (s - a) . . . .$ $L [2 e^{3 t} s i n (4 t)] = 2 * \frac{4}{{(s - 3)}^{2} + 16}$ $= \frac{8}{s^{2} - 6 s + 25}$
	Commented by BHOOPENDRA last updated on 07/Jan/21

	$t s i n h 2 t s i n 3 t$ $p l z h e l p m e o u t t h i s a l s o$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum