2. Find the Inverse Laplace Transform of following...

2. Find the Inverse Laplace Transform of following functions. (a) (5 points) \(F(s) = \frac{e^{-2s}}{(s+3)^4}\) (b) (8 points) \(F(s) = \frac{s-3}{s^2+4s+6}\) (c) (12 points) \(F(s) = \frac{s}{(s+1)(s^2-2s+1)}\)

Transcript

00:01 To find the inverse laplace transform of the following functions.

00:06 Okay, so first function given as f of s is equal to 2 upon s raised to power 4 plus 2 upon twice s plus 1 plus 2s upon s square plus 16.

00:21 We have to find the inverse laplace transform of this function.

00:24 So, laplace inverse of f of s is equal to laplace inverse of 2 upon s raised to power 4 plus 2 upon twice s plus 1 plus twice s upon s square plus 16.

00:44 Okay, now here we use the linearity property of inverse laplace transform.

00:49 That is laplace transform of a into f of s plus b into g of s, it will be equal to a into laplace inverse of f of s plus b into laplace inverse of g of s.

01:06 This property here we have to use.

01:08 So this implies laplace inverse of f of s, it is equal to laplace inverse of 2 upon s raised to power 4 plus laplace inverse of 2 upon twice s plus 1 plus laplace inverse of twice s upon s square plus 16.

01:31 Now laplace inverse of 2 upon s raised to power 4 is nothing but equal to t cube upon 3 and laplace inverse of 2 upon twice s plus 1 is nothing but e raised to minus t upon 2.

01:46 Then laplace inverse of twice s upon s square plus 16 is nothing but equal to 2 into cos of 4t.

01:55 So, this is the required laplace transform for this first function f of s.

02:02 This is the required answer.

02:06 That is laplace inverse of f of s is equal to t cube upon 3 plus e raised to minus t by 2 plus 2 cos of 4t.

02:15 Now in b question, we have to find the laplace inverse of 2 upon s minus 3 raised to power 4 plus twice s upon s square plus 4s plus 20.

02:30 We have to calculate this laplace inverse.

02:33 Now here we expand this term twice s upon s square plus 4s plus 20.

02:39 Its expansion is 2 into s plus 2 divided by s plus 2 whole square plus 16 minus 4 into 1 upon s plus 2 square plus 16.

02:59 Now this will implies that laplace inverse will be equal to laplace inverse of 2 upon s minus 3 raised to power 4 as it is and this term is replaced by this two numbers that is 2 into s plus 2 divided by s plus 2 square plus 16 minus 4 into 1 upon s plus 2 square plus 16.

03:27 Then we use the linearity property of inverse laplace transform that is laplace inverse of a into f of s plus b into g of s.

03:38 This will be equal to the laplace inverse a into laplace inverse of f of s plus b into laplace inverse of g of s...

Question

2. Find the Inverse Laplace Transform of following functions. (a) (5 points) \(F(s) = \frac{e^{-2s}}{(s+3)^4}\) (b) (8 points) \(F(s) = \frac{s-3}{s^2+4s+6}\) (c) (12 points) \(F(s) = \frac{s}{(s+1)(s^2-2s+1)}\)

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