What is inverse oflaplace (5+5s)/(s^2+25) ?

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

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inverse oflaplace (5+5s)/(s^2+25)

inverse laplace

\frac{5 + 5 s}{s ^{2} + 25}

sin (5 t) + 5 cos (5 t)

Solution steps

L^{- 1} {\frac{5 + 5 s}{s ^{2} + 25}}

Expand

= L^{- 1} {\frac{5}{s ^{2} + 25} + \frac{5 s}{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)}

= 5 L^{- 1} {\frac{1}{s ^{2} + 25}} + 5 L^{- 1} {\frac{s}{s ^{2} + 25}}

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

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

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

Refine

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

= sin (5 t) + 5 cos (5 t)

x^{4} + 2 e^{x}

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\int_{0}^{2} 1 + x d x

\frac{x}{1 - \frac{1}{x ^{2}}}

\int 3 (3 x + 9)^{5} d x

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