Question

inverse laplace of L^-1 {5s/s^2 + 3s - 4}​

Answer 1

Answer:

$L^{-1} (\frac{5s}{s^{2}+ 3 s -4 } )$ =  $4e^{-4t} + e^{t}$

Step-by-step explanation:

Step(i):-

Given

           $L^{-1} (\frac{5s}{s^{2}+ 3 s -4 } )$

Factors of    s² + 3s - 4

               =    s² + 4s - s -4

               =   s( s +4 ) -1 (s +4)

               =   (s-1)(s+4)

  $L^{-1} (\frac{5s}{s^{2}+ 3 s -4 } )$

              =    $L^{-1} (\frac{5s}{s+4)(s-1) } )$

By using partial fractions

           $(\frac{s}{s+4)(s-1) } ) = \frac{A}{s+4} + \frac{B}{s-1}$ ..(i)

                s  = A ( s-1) + B( s+4) ....(ii)

          Put s= 1  in equation (ii) , we get

             1 = B(5)

           $B = \frac{1}{5}$

         s = -4 in equation (ii) , we get

        -4 = -5A

         $A =\frac{4}{5}$

Step(ii):-

    now the equation (i) , we get

$(\frac{s}{s+4)(s-1) } ) = \frac{4}{5(s+4)} + \frac{1}{5(s-1)}$

$L^{-1} (\frac{s}{s+4)(s-1) } ) = 5( L^{-1} \frac{4}{5(s+4)} + 5L^{-1} \frac{1}{5(s-1)}$                      

By using inverse Laplace transform formula

$L^{-1} (\frac{1}{s-a} ) = e^{at}$

$L^{-1} (\frac{1}{s-1} ) = e^{t}$

$L^{-1} (\frac{1}{s+4} ) = e^{-4t}$

$L^{-1} (\frac{s}{s+4)(s-1) } ) = 5( L^{-1} \frac{4}{5(s+4)} + 5L^{-1} \frac{1}{5(s-1)}$                    

                      = $4e^{-4t} + e^{t}$