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)

\frac{s ^{2}}{s ^{2} + 4 s + 5}

4 - x^{2}, 14 - 7 x

x^{3} - 3 x^{2} + 3 x - 3

(7 x^{\frac{1}{5}}) - (6 x^{\frac{4}{7}})

\frac{d}{d x} (5 2 x + 6)

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