Inverse laplace of [(2s^2 -4s + 5)/s^3] = ?
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Answer:
What is the inverse Laplace of (4s-3) / (s^2-4S-5)?
(7/6)e^-t +(17/6)e^[5t]
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(4s -3)/(s^2 -4s -5) = (4s-3)/[(s +1)(s-5)]
do a partial fraction expansion
(4s-3) = A/(s +1) + B/(s -5)
4s -3 = As -5A + Bs + B
A + B = 4
-5A + B = -3
6A =7
A = 7/6
B = 17/6
(4s-3)/[(s+1)(s -5)] = (7/6)/(s + 1) + (17/6)/(s-5)
(7/6) e^-t +( 17/6) e^5t
Step-by-step explanation:
What is the inverse Laplace of (4s-3) / (s^2-4S-5)?
(7/6)e^-t +(17/6)e^[5t]
details
(4s -3)/(s^2 -4s -5) = (4s-3)/[(s +1)(s-5)]
do a partial fraction expansion
(4s-3) = A/(s +1) + B/(s -5)
4s -3 = As -5A + Bs + B
A + B = 4
-5A + B = -3
6A =7
A = 7/6
B = 17/6
(4s-3)/[(s+1)(s -5)] = (7/6)/(s + 1) + (17/6)/(s-5)
(7/6) e^-t +( 17/6) e^5t
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