Question

inverse laplace of (s²+8s+27)/(s+1)(s²+4s+13)​

Answer 1

Step-by-step explanation:

f(t) = L−1{F}(t). Table of inverse Laplace

transform. F(s) f(t) = L−1{F}(t). 1 ... Example 2.

Determine L−1 { 3. (2s+5)3 + 2s+16 s2+4s+13.

+. 3 s2+4s+8}. ... A( s + 2)(s + 5) + B(s − 1)(s + 5)

+ C(s − 1)(s + 2).

#khushu

Answer 2

Answer:

Inverse Laplace of (s²+8s+27)/(s+1)(s²+4s+13)​ is  2e^(-t) - e^(-2t)cos(3t) + e^(-2t)sin(3t)

Step-by-step explanation:

Let F(s) = (s²+8s+27)/(s+1)(s²+4s+13)​

Step 1 : Reducing to partial fraction

F(s) =  $2/(1+s) + (1-s)/(s^2+4s +13)$

F(s) = $2/(1+s) - (s+2)/(s^2+4s +13)+3/(s^2+4s +13)$

F(s) = $2/(1+s) - (s+2)/((s+2)^2+9) + 3/((s+2)^2+9)$

# Inverse Laplace of      $\frac{1}{s+a} = e^{-at}$

# Inverse Laplace of     $\frac{s}{s^{2}+w^{2} } = cos(wt)$

# Inverse Laplace of     $\frac{w}{s^{2}+w^{2} } = sin(wt)$

If Laplace transform of f(t) is F(s) then,

 Laplace transform of  $e^{-at} f(t)$ = F(s+a)

Therefore,

If Inverse Laplace of F(s) is f(t), then inverse laplace of F(s+a) is $e^{-at} f(t)$  

Step 2 : Taking inverse laplace transform on both sides

f(t) =  $2e^(^-^t^) - e^(^-^2^t^)cos(3t) + e^(^-^2^t^)sin(3t)$

#SPJ3