2e-3t-sin4t-find-laplace-transformation-

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) \dots .

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