IDNLearn.com offers a comprehensive solution for finding accurate answers quickly. Join our community to receive prompt and reliable responses to your questions from knowledgeable professionals.
Sagot :
Answer:
[tex]L^{-1} (\frac{5s}{s^{2}+ 3 s -4 } )[/tex] = [tex]4e^{-4t} + e^{t}[/tex]
Step-by-step explanation:
Step(i):-
Given
[tex]L^{-1} (\frac{5s}{s^{2}+ 3 s -4 } )[/tex]
Factors of s² + 3s - 4
= s² + 4s - s -4
= s( s +4 ) -1 (s +4)
= (s-1)(s+4)
[tex]L^{-1} (\frac{5s}{s^{2}+ 3 s -4 } )[/tex]
= [tex]L^{-1} (\frac{5s}{s+4)(s-1) } )[/tex]
By using partial fractions
[tex](\frac{s}{s+4)(s-1) } ) = \frac{A}{s+4} + \frac{B}{s-1}[/tex] ..(i)
s = A ( s-1) + B( s+4) ....(ii)
Put s= 1 in equation (ii) , we get
1 = B(5)
[tex]B = \frac{1}{5}[/tex]
s = -4 in equation (ii) , we get
-4 = -5A
[tex]A =\frac{4}{5}[/tex]
Step(ii):-
now the equation (i) , we get
[tex](\frac{s}{s+4)(s-1) } ) = \frac{4}{5(s+4)} + \frac{1}{5(s-1)}[/tex]
[tex]L^{-1} (\frac{s}{s+4)(s-1) } ) = 5( L^{-1} \frac{4}{5(s+4)} + 5L^{-1} \frac{1}{5(s-1)}[/tex]
By using inverse Laplace transform formula
[tex]L^{-1} (\frac{1}{s-a} ) = e^{at}[/tex]
[tex]L^{-1} (\frac{1}{s-1} ) = e^{t}[/tex]
[tex]L^{-1} (\frac{1}{s+4} ) = e^{-4t}[/tex]
[tex]L^{-1} (\frac{s}{s+4)(s-1) } ) = 5( L^{-1} \frac{4}{5(s+4)} + 5L^{-1} \frac{1}{5(s-1)}[/tex]
= [tex]4e^{-4t} + e^{t}[/tex]
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
