What is inverse oflaplace (s+1)/(s^2+25) ?

The answer to inverse oflaplace (s+1)/(s^2+25) is cos(5t)+1/5 sin(5t)

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inverse oflaplace (s+1)/(s^2+25)

inverse laplace

\frac{s + 1}{s ^{2} + 25}

cos (5 t) + \frac{1}{5} sin (5 t)

Solution steps

L^{- 1} {\frac{s + 1}{s ^{2} + 25}}

Expand

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

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

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

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

= cos (5 t) + \frac{1}{5} sin (5 t)

\int 10 x e^{x^{2}} d x

f (x) = 1 + 2 e^{3 x}

8 (x - \frac{1}{x})^{2}

\frac{d ^{2} y}{d x ^{2}} y + 81 y = 0

y^{^{'}} = k x

What is inverse oflaplace (s+1)/(s^2+25) ?
The answer to inverse oflaplace (s+1)/(s^2+25) is cos(5t)+1/5 sin(5t)