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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]
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