What is inverse oflaplace s/(s^2+10s+25) ?

The answer to inverse oflaplace s/(s^2+10s+25) is e^{-5t}-e^{-5t}*5t

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inverse oflaplace s/(s^2+10s+25)

inverse laplace

\frac{s}{s ^{2} + 10 s + 25}

e^{- 5 t} - e^{- 5 t} \cdot 5 t

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 10 s + 25}}

Take the partial fraction of

\frac{s}{s ^{2} + 10 s + 25} : \frac{1}{s + 5} - \frac{5}{( s + 5 ) ^{2}}

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

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

L^{- 1} {\frac{1}{s + 5}} : e^{- 5 t}

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

= e^{- 5 t} - e^{- 5 t} \cdot 5 t

x \to 0 lim (\frac{5}{x} - \frac{5}{x ^{2} + x})

t \to 1 lim (t + 3)

f (x) = 2 x^{2} - 5 x + 3

x \to 0 + lim (ln (1 + x))

y = 5 tan (x), (\frac{π}{4}, 5)

What is inverse oflaplace s/(s^2+10s+25) ?
The answer to inverse oflaplace s/(s^2+10s+25) is e^{-5t}-e^{-5t}*5t