Help please with this Laplace Inverse Transformation.?

s/(s^4 + 4a^4)

= s/[(s^4 + 4a^2 s^2 + 4a^4) - 4a^2 s^2], completing the square

= s/[(s^2 + 2a^2)^2- (2as)^2]

= s/[(s^2 - 2as + 2a^2) (s^2 + 2as + 2a^2)]

= (1/(4a)) * [1/(s^2 - 2as + 2a^2) - 1/(s^2 + 2as + 2a^2)], via partial fractions

= (1/(4a)) * [1/((s - a)^2 + a^2) - 1/((s + a)^2 + a^2)], more completing the square.

Inverting now yields

(1/(4a)) * [(1/a) e^(at) sin(at) - (1/a) e^(-at) sin(at)]

= (1/(2a^2)) * sin(at) * [e^(at) - e^(-at)]/2

= (1/(2a^2)) sin(at) sinh(at).

I hope this helps!

Laplace Inverse Of S

Partial fractions :) The denominator facotrs to : 2/[(s)(s+2)(s+4) = a/s + b/(s+2) + c/(s+4). a(s+2)(s+4) + b(s)(s+4) + c(s)(s+2) = as^2 + 6a*s + 8a + bs^2 + 4bs +cs^2 + 2cs. =2. The term without 's' is "8a" it truly is as a result equivalent to 2. a = one million/4. (a+b+c) [the coefficients of s^2] =0. 6a+4b+2c [the coefficients of s] = 0. b+c = -one million/4. 2b+2c= -one million/2 4b+2c= -3/2. -2b = one million b=-one million/2. c=one million/4. = (one million/4)/s + (-one million/2)/(s+2) + (one million/4)/(s+4). Inverse laplacing gets you: one million/4 -one million/2*e^(-2t) +one million/4*e^(-4t)