What is inverse oflaplace 7/(s^2+5s+10) ?

The answer to inverse oflaplace 7/(s^2+5s+10) is (14}{sqrt(15)}e^{-\frac{5t)/2}sin((sqrt(15)t)/2)

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inverse oflaplace 7/(s^2+5s+10)

inverse laplace

\frac{7}{s ^{2} + 5 s + 10}

\frac{14}{15} e^{- \frac{5 t}{2}} sin (\frac{15 t}{2})

Solution steps

L^{- 1} {\frac{7}{s ^{2} + 5 s + 10}}

Expand

\frac{7}{s ^{2} + 5 s + 10} : 7 \cdot \frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{15}{4}}

= L^{- 1} {7 \cdot \frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{15}{4}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 7 L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{15}{4}}}

L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{15}{4}}} : e^{- \frac{5 t}{2}} \frac{2}{15} sin (\frac{15 t}{2})

= 7 e^{- \frac{5 t}{2}} \frac{2}{15} sin (\frac{15 t}{2})

Refine

7 e^{- \frac{5 t}{2}} \frac{2}{15} sin (\frac{15 t}{2}) : \frac{14}{15} e^{- \frac{5 t}{2}} sin (\frac{15 t}{2})

= \frac{14}{15} e^{- \frac{5 t}{2}} sin (\frac{15 t}{2})

(20 + x^{2}) y^{^{'}} - 2 x y = 0

\frac{d}{d x} ((x^{2} + 1))

(4 x - 3)^{4}

f (x) = (- 9 x^{2} + 4)^{2}

\int x arctan (7 x) d x

What is inverse oflaplace 7/(s^2+5s+10) ?
The answer to inverse oflaplace 7/(s^2+5s+10) is (14}{sqrt(15)}e^{-\frac{5t)/2}sin((sqrt(15)t)/2)