What is inverse oflaplace 1/(s^2+5s+2) ?

The answer to inverse oflaplace 1/(s^2+5s+2) is e^{-(5t)/2} 2/(sqrt(17))sinh((sqrt(17)t)/2)

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inverse oflaplace 1/(s^2+5s+2)

inverse laplace

\frac{1}{s ^{2} + 5 s + 2}

e^{- \frac{5 t}{2}} \frac{2}{17} sinh (\frac{17 t}{2})

Solution steps

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

\frac{1}{s ^{2} + 5 s + 2} = \frac{1}{( s + \frac{5}{2} ) ^{2} - \frac{17}{4}}

= L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} - \frac{17}{4}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s + \frac{5}{2} ) ^{2} - \frac{17}{4}} : a = - \frac{5}{2}, F (s) = \frac{1}{s ^{2} - \frac{17}{4}}

= e^{- \frac{5}{2} t} L^{- 1} {\frac{1}{s ^{2} - \frac{17}{4}}}

L^{- 1} {\frac{1}{s ^{2} - \frac{17}{4}}} : \frac{2}{17} sinh (\frac{17 t}{2})

= e^{- \frac{5 t}{2}} \frac{2}{17} sinh (\frac{17 t}{2})

(\frac{3 x + 2}{x - 1})^{7}

(2 x^{3} + 7)^{2}

\int x^{2} \cdot 4^{- x} d x

f (x) = x^{2}, (0, - 25)

\int 2 x^{2} + 2 d x

What is inverse oflaplace 1/(s^2+5s+2) ?
The answer to inverse oflaplace 1/(s^2+5s+2) is e^{-(5t)/2} 2/(sqrt(17))sinh((sqrt(17)t)/2)