What is inverse oflaplace 3/((s^2+4s+13)) ?

The answer to inverse oflaplace 3/((s^2+4s+13)) is e^{-2t}sin(3t)

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inverse oflaplace 3/((s^2+4s+13))

inverse laplace

\frac{3}{( s ^{2} + 4 s + 13 )}

e^{- 2 t} sin (3 t)

Solution steps

L^{- 1} {\frac{3}{( s ^{2} + 4 s + 13 )}}

Expand

\frac{3}{s ^{2} + 4 s + 13} : 3 \cdot \frac{1}{( s + 2 ) ^{2} + 9}

= L^{- 1} {3 \cdot \frac{1}{( s + 2 ) ^{2} + 9}}

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)}

= 3 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 9}}

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 9}} : e^{- 2 t} \frac{1}{3} sin (3 t)

= 3 e^{- 2 t} \frac{1}{3} sin (3 t)

Refine

3 e^{- 2 t} \frac{1}{3} sin (3 t) : e^{- 2 t} sin (3 t)

= e^{- 2 t} sin (3 t)

\frac{d y}{d x} - (sin (x)) y = 2 sin (x)

f (x) = cos^{2} (x) - 2 sin (x)

\int_{0}^{\infty} cos (π x) d x

\int 8 cos^{2} (x) d x

(1 x y^{2} + 4 y) d x + (1 x^{2} y + 4 x) d y = 0

What is inverse oflaplace 3/((s^2+4s+13)) ?
The answer to inverse oflaplace 3/((s^2+4s+13)) is e^{-2t}sin(3t)