Find the inverse of Laplace transform f(s)=7s+14/s²+2

Question

karton · Accepted Answer

To find the inverse Laplace transform of

F (s) = \frac{7 s + 14}{s^{2} + 2}

, we can use partial fraction decomposition and known Laplace transforms.
First, let's factor the denominator of

F (s)

:

s^{2} + 2 = (s + \sqrt{2}) (s - \sqrt{2})

Now, we can express

F (s)

as partial fractions:

F (s) = \frac{A}{s + \sqrt{2}} + \frac{B}{s - \sqrt{2}}

To find the values of

A

and

B

, we can multiply both sides of the equation by the denominator

(s^{2} + 2)

and simplify:

(7 s + 14) = A (s - \sqrt{2}) + B (s + \sqrt{2})

Expanding and collecting like terms:

7 s + 14 = (A + B) s + (B \sqrt{2} - A \sqrt{2})

Comparing the coefficients of like powers of

s

:

A + B = 7

B \sqrt{2} - A \sqrt{2} = 14

Solving these equations, we find

A = 3

and

B = 4

.
Now, we can rewrite

F (s)

in terms of the partial fractions:

F (s) = \frac{3}{s + \sqrt{2}} + \frac{4}{s - \sqrt{2}}

Taking the inverse Laplace transform of each term using known transforms:

ℒ^{- 1} {\frac{3}{s + \sqrt{2}}} = 3 e^{- \sqrt{2} t}

ℒ^{- 1} {\frac{4}{s - \sqrt{2}}} = 4 e^{\sqrt{2} t}

Therefore, the inverse Laplace transform of

F (s) = \frac{7 s + 14}{s^{2} + 2}

is:

ℒ^{- 1} {F (s)} = 3 e^{- \sqrt{2} t} + 4 e^{\sqrt{2} t}

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