What is inverse oflaplace (7s+53)/(s^2+10s+61) ?

The answer to inverse oflaplace (7s+53)/(s^2+10s+61) is 7e^{-5t}cos(6t)+3e^{-5t}sin(6t)

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inverse oflaplace (7s+53)/(s^2+10s+61)

inverse laplace

\frac{7 s + 53}{s ^{2} + 10 s + 61}

7 e^{- 5 t} cos (6 t) + 3 e^{- 5 t} sin (6 t)

Solution steps

L^{- 1} {\frac{7 s + 53}{s ^{2} + 10 s + 61}}

Expand

\frac{7 s + 53}{s ^{2} + 10 s + 61} : 7 \cdot \frac{s + 5}{( s + 5 ) ^{2} + 36} + 18 \cdot \frac{1}{( s + 5 ) ^{2} + 36}

= L^{- 1} {7 \cdot \frac{s + 5}{( s + 5 ) ^{2} + 36} + 18 \cdot \frac{1}{( s + 5 ) ^{2} + 36}}

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)}

= 7 L^{- 1} {\frac{s + 5}{( s + 5 ) ^{2} + 36}} + 18 L^{- 1} {\frac{1}{( s + 5 ) ^{2} + 36}}

L^{- 1} {\frac{s + 5}{( s + 5 ) ^{2} + 36}} : e^{- 5 t} cos (6 t)

L^{- 1} {\frac{1}{( s + 5 ) ^{2} + 36}} : e^{- 5 t} \frac{1}{6} sin (6 t)

= 7 e^{- 5 t} cos (6 t) + 18 e^{- 5 t} \frac{1}{6} sin (6 t)

Refine

7 e^{- 5 t} cos (6 t) + 18 e^{- 5 t} \frac{1}{6} sin (6 t) : 7 e^{- 5 t} cos (6 t) + 3 e^{- 5 t} sin (6 t)

= 7 e^{- 5 t} cos (6 t) + 3 e^{- 5 t} sin (6 t)

\int cos (2 x) cos (3 x) d x

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

\frac{d}{d x} (ln (x^{2} - 6 x + 1))

y = 4 x, y = 4 x

\int_{0}^{0.5} sin (t^{2}) d t

What is inverse oflaplace (7s+53)/(s^2+10s+61) ?
The answer to inverse oflaplace (7s+53)/(s^2+10s+61) is 7e^{-5t}cos(6t)+3e^{-5t}sin(6t)