What is inverse oflaplace ((4s-5))/(s^2-6s+13) ?

The answer to inverse oflaplace ((4s-5))/(s^2-6s+13) is 4e^{3t}cos(2t)+7/2 e^{3t}sin(2t)

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inverse oflaplace ((4s-5))/(s^2-6s+13)

inverse laplace

\frac{( 4 s - 5 )}{s ^{2} - 6 s + 13}

4 e^{3 t} cos (2 t) + \frac{7}{2} e^{3 t} sin (2 t)

Solution steps

L^{- 1} {\frac{( 4 s - 5 )}{s ^{2} - 6 s + 13}}

Expand

\frac{4 s - 5}{s ^{2} - 6 s + 13} : 4 \cdot \frac{s - 3}{( s - 3 ) ^{2} + 4} + 7 \cdot \frac{1}{( s - 3 ) ^{2} + 4}

= L^{- 1} {4 \cdot \frac{s - 3}{( s - 3 ) ^{2} + 4} + 7 \cdot \frac{1}{( s - 3 ) ^{2} + 4}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= 4 L^{- 1} {\frac{s - 3}{( s - 3 ) ^{2} + 4}} + 7 L^{- 1} {\frac{1}{( s - 3 ) ^{2} + 4}}

L^{- 1} {\frac{s - 3}{( s - 3 ) ^{2} + 4}} : e^{3 t} cos (2 t)

L^{- 1} {\frac{1}{( s - 3 ) ^{2} + 4}} : e^{3 t} \frac{1}{2} sin (2 t)

= 4 e^{3 t} cos (2 t) + 7 e^{3 t} \frac{1}{2} sin (2 t)

Refine

4 e^{3 t} cos (2 t) + 7 e^{3 t} \frac{1}{2} sin (2 t) : 4 e^{3 t} cos (2 t) + \frac{7}{2} e^{3 t} sin (2 t)

= 4 e^{3 t} cos (2 t) + \frac{7}{2} e^{3 t} sin (2 t)

n = 1 \sum \infty (\frac{1}{3})^{n}

4 x^{2}, \frac{x}{4}

y = x^{2} + 4, y = - x^{2} - 1, x = - 2, x = 3

30 ln (2 x + 1)

x \to - 3 - lim (ln (x^{2} - 9))

What is inverse oflaplace ((4s-5))/(s^2-6s+13) ?
The answer to inverse oflaplace ((4s-5))/(s^2-6s+13) is 4e^{3t}cos(2t)+7/2 e^{3t}sin(2t)