Plese help! Find the inverse laplace of (4-s)/((s^2)+4)(s+1) I know the answer i
ID: 1808296 • Letter: P
Question
Plese help! Find the inverse laplace of (4-s)/((s^2)+4)(s+1) I know the answer is exp(-t) - cos(2*t), and that the answer comes from 1/(s + 1) - s/(s^2 + 4) But I dont know how to get from (4-s)/((s^2)+4)(s+1) to 1/(s + 1) - s/(s^2 + 4)Explanation / Answer
(4-s)/((s^2)+4)(s+1) => (A/ s+1 )+ ((B s+C)/s^2+4) To find value of A,B,C take l c m add both numerator , make it equal to (4-s) A(s^2+4) + (B s+C)(s+1) = 4-s As^2 + 4A + B s^2 + Cs + Bs+ C=4-s (A+B)*s^2+ (B+C)s+ 4A+C =4-s so from above A+B= 0 (A=-B) B+C= -1 (=> -A+C=-1) 4A+C=4 ( 4A+C=4) so A= 1 , C=0 and B=-A=-1 so put this value in A/ s+1 )+ ((B s+C)/s^2+4) => 1/s+1 - s/s^2+4