use the table of laplace transforms at the end of this chapter to compute the inverse laplace transform of the following function
a. 2s/s(s的二次方-3s+2)
please steps by steps!
2s/[s(s^2-3s+2)]
=2/(s^2-3s+2)
=2/[(s-1)(s-2)]
= 2/(s-2) - 2/(s-1)
so that,
the inverse Laplace transform of 2s/[s(s^2-3s+2)]
= L^(-1) { 2/(s-2) - 2/(s-1) }
= 2 e^(2x) - 2 e^x
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Verified answer
2s/[s(s^2-3s+2)]
=2/(s^2-3s+2)
=2/[(s-1)(s-2)]
= 2/(s-2) - 2/(s-1)
so that,
the inverse Laplace transform of 2s/[s(s^2-3s+2)]
= L^(-1) { 2/(s-2) - 2/(s-1) }
= 2 e^(2x) - 2 e^x