Question

find the inverse laplace transform of
14S+10/49S²+28S+13​

Answer 1

The inverse Laplace transformation of the equation is 2cos(3t) + 2sin(3t)

GIVEN:-

$14S+10/49S²+28S+13$

TO FIND:- inverse Laplace transform of the given

SOLUTION:-

${L}^{ - 1} \times \frac{(14s + 10)}{(49s ^ 2 + 28s + 13)}$

${L}^{ - 1} \times (14s + 10)/((7s + 2) ^ 2 + 3 ^ 2)$

${L}^{ - 1} \times (2(7s + 2) + 6)/((7s + 2) ^ 2 + 3 ^ 2)$

${2L}^{ - 1} \times (7s + 2)/((7s + 2) ^ 2 + (3) ^ 2) + {6L}^{ - 1} \times 1/((7s + 2) ^ 2 + (3) ^ 2)$

$2 cos(3t)+ 6 \times 1/3 \times sin(3t)$

$2cos(3t) + 2sin(3t)$

Hence, The inverse Laplace transformation of the equation is 2cos(3t) + 2sin(3t)

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