What is inverse oflaplace ((2s+3))/(s^2+2s+26) ?

The answer to inverse oflaplace ((2s+3))/(s^2+2s+26) is 2e^{-t}cos(5t)+1/5 e^{-t}sin(5t)

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inverse oflaplace ((2s+3))/(s^2+2s+26)

inverse laplace

\frac{( 2 s + 3 )}{s ^{2} + 2 s + 26}

2 e^{- t} cos (5 t) + \frac{1}{5} e^{- t} sin (5 t)

Solution steps

L^{- 1} {\frac{( 2 s + 3 )}{s ^{2} + 2 s + 26}}

Expand

\frac{2 s + 3}{s ^{2} + 2 s + 26} : 2 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 25} + 1 \cdot \frac{1}{( s + 1 ) ^{2} + 25}

= L^{- 1} {2 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 25} + 1 \cdot \frac{1}{( s + 1 ) ^{2} + 25}}

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)}

= 2 L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 25}} + 1 \cdot L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 25}}

L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 25}} : e^{- t} cos (5 t)

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 25}} : e^{- t} \frac{1}{5} sin (5 t)

= 2 e^{- t} cos (5 t) + 1 \cdot e^{- t} \frac{1}{5} sin (5 t)

Refine

2 e^{- t} cos (5 t) + 1 e^{- t} \frac{1}{5} sin (5 t) : 2 e^{- t} cos (5 t) + \frac{1}{5} e^{- t} sin (5 t)

= 2 e^{- t} cos (5 t) + \frac{1}{5} e^{- t} sin (5 t)

\int \frac{6 x}{1 - 7 x ^{2}} d x

2 x y^{^{'}} + y = 10 x

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

\int (x^{3} + 1)^{3} d x

What is inverse oflaplace ((2s+3))/(s^2+2s+26) ?
The answer to inverse oflaplace ((2s+3))/(s^2+2s+26) is 2e^{-t}cos(5t)+1/5 e^{-t}sin(5t)