Question

Find
$l {}^{ - 1} (s \div s {}^{2} + b {}^{2} )$
​

Answer 1

Answer:

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Answer 2

l^-1(s/s^2 + b^2)
= l^-1 [(s^1/s^2)+(b^2)]

a^-m = 1/a^m
a^m/a^m = a^m-n

= 1/l [s^(1-2) + b^2]
= 1/l [s^-1 + b^2]
= 1/l [1/s + b^2]
= [(1/l)(1/s)+(1/l)(b^2)]
= 1/ls + b^2/l

Therefore the value for the following Question is 1/ls + b^2/l

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