Respuesta :

Answer:

[tex]L^{-1} \begin {pmatrix} \dfrac{s}{((s-20)(s-5)(s-4)} \end {pmatrix}= \dfrac{e^{4t}}{12}\begin {pmatrix}e^{16t} - 4e^t+3 \end {pmatrix}[/tex]

Step-by-step explanation:

From the given information the first step to carry out is to find the partial fraction of the given equation; i.e.

[tex]\dfrac{s}{(s-4)(s-5)(s-20)}= \dfrac{A}{s-4}+\dfrac{B}{s-5}+\dfrac{C}{s-20}[/tex]

After finding the partial fraction; we get:

[tex]s= s^2(A+B+C)+s(-2A-24B-25C)+20A+80B+100C[/tex]

By solving this; we get:

[tex]A =\dfrac{1}{12}\ ; \ B = -\dfrac{1}{3} \ ; \ C= \dfrac{1}{4}[/tex]

Thus;  the above equation can be re-written as:

[tex]\dfrac{s}{(s-20)(s-5)(s-4)}= \dfrac{\dfrac{1}{12}}{s-20}+\dfrac{-\dfrac{1}{3} }{s-5}+\dfrac{\dfrac{1}{4}}{s-4}[/tex]

Finally as a function of t ;

[tex]L^{-1} \begin {pmatrix} \dfrac{s}{((s-20)(s-5)(s-4)} \end {pmatrix}= \dfrac{e^{4t}}{12}\begin {pmatrix}e^{16t} - 4e^t+3 \end {pmatrix}[/tex]

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