Find the inverse Laplace transform ft of the function Fs 10

Find the inverse Laplace transform f(t) of the function F(s) = 10s + 3/25 - s^2.

Solution

Given that

F(s) = ( 10s + 3 ) / ( 25 - s² )

Take

( 10s + 3 ) / ( 25 - s² ) into partial fractions

( 10s + 3 ) / ( 25 - s² ) = ( 10s + 3 ) / ( 5 + s ) ( 5 - s )

= A / ( 5 + s ) + B / ( 5 - s )

= [ A(5 - s ) + B( 5 + s ) ] / [ ( 5 + s ) ( 5 - s ) ]

( 10s + 3 ) / ( 25 - s² ) = [ 5A - As + 5B + Bs ] / [ ( 5 + s ) ( 5 - s ) ]

10s + 3 = 5A - As + 5B + Bs

10s + 3 = s( -A + B ) + 5A + 5B

Equating the coefficients

-A + B = 10

5A + 5B = 3

On solving these tw equations

A = -47/10 , B = 53/10

Hence,

  ( 10s + 3 ) / ( 25 - s² ) = A / ( 5 + s ) + B / ( 5 - s )

( 10s + 3 ) / ( 25 - s² ) = (-47/10)/(5 + s) + (53/10) / (5 - s)

= -1 [ (47/10) /(s + 5) + (53/10) / (s - 5) ]

= (-47/10) ( 1/(s+5) ) - (53/10) ( 1/ (s - 5 ) )

f(t) = L^-1 {  ( 10s + 3 ) / ( 25 - s² ) } = L^-1 { (-47/10) (1/(s+5)) - (53/10) ( 1/ (s - 5 ) ) }

= L^-1 { (-47/10) (1/(s+5)) } - L^-1 { (53/10) ( 1/ (s - 5 ) ) }

= (-47/10) L^-1(1/(s+5)) - (53/10) L^-1( 1/ (s - 5 ) )

= (-47/10) .e^-5t + (-53/10)e^5t [since L^-1( 1/s-a) = e^at]

Therefore,

f(t) = (-47/10)e^-5t + (-53/10)e^5t