Question

find the inverse laplace of 1/(s(s-3))

Answer 1

\[ {1 \over s (s-3)} = {1 \over -3}\left ( {1 \over s} - {1 \over s- 3} \right)\]
take inverse Laplace transform.

Answer 2

what did you do?

Answer 3

\[\frac{ 1 }{ 3 }e ^{3t}\]

Answer 4

okay so you split it up right?

Answer 5

yep!! http://en.wikipedia.org/wiki/Heaviside_cover-up_method

Answer 6

i dont understand so 1/s + 1/(s-3)

Answer 7

\[\frac{ 1 }{ 3 }(1-e ^{3t})\]

Answer 8

i dont need the answer haha

Answer 9

And giving answer is against the Code of Conduct of this site!

Answer 10

You get it in a form so that you can use the laplace formulas.

Answer 11

okay so can i use partial fractions

Answer 12

like 1/3(s-3) - 1/3s

Answer 13

Partial fractions \[\frac1{s(s-3)}=\frac{A}{s}+\frac{B}{s-3}\]

\[1=A(s-3)+B(s)\]

when \(s=0\)
\[1=-3A\qquad \implies\qquad A=-\frac13\]

when \(s=3\)
\[1=3B\qquad \implies\qquad B=\frac13\]

\[\frac1{s(s-3)}=\frac{-1/3}{s}+\frac{1/3}{s-3}=\frac13\left(\frac1{s-3}-\frac{1}{s}\right)\]

Answer 14

okay where do i go from there?

Answer 15

1/s is just t

Answer 16

i meant 1

Answer 17

now take the inverse laplace transform

Answer 18

how do i take it for 1/3(s-3)

Answer 19

oh its just e^455465

Answer 20

\[\mathcal L\Big\{\frac13\left(\frac1{s-3}-\frac{1}{s}\right)\Big\}\]
\[=\frac13\left(\mathcal L\Big\{\frac1{s-3}\Big\}-\mathcal L\Big\{\frac{1}{s}\Big\}\right)\]

Answer 21

1/3 * e^3t - 1/3

Answer 22

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