What is inverse oflaplace x/((x+2)(x^2+4)) ?

The answer to inverse oflaplace x/((x+2)(x^2+4)) is -1/4 e^{-2t}+1/4 cos(2t)+(sin(2t))/4

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inverse oflaplace x/((x+2)(x^2+4))

inverse laplace

\frac{x}{( x + 2 ) ( x ^{2} + 4 )}

- \frac{1}{4} e^{- 2 t} + \frac{1}{4} cos (2 t) + \frac{sin ( 2 t )}{4}

Solution steps

L^{- 1} {\frac{x}{( x + 2 ) ( x ^{2} + 4 )}}

Take the partial fraction of

\frac{x}{( x + 2 ) ( x ^{2} + 4 )} : - \frac{1}{4 ( x + 2 )} + \frac{x + 2}{4 ( x ^{2} + 4 )}

= L^{- 1} {- \frac{1}{4 ( x + 2 )} + \frac{x + 2}{4 ( x ^{2} + 4 )}}

Expand

= L^{- 1} {\frac{x}{4 ( x ^{2} + 4 )} + \frac{2}{4 ( x ^{2} + 4 )} - \frac{1}{4 ( x + 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)}

= - \frac{1}{4} L^{- 1} {\frac{1}{x + 2}} + \frac{1}{4} L^{- 1} {\frac{x}{x ^{2} + 4}} + \frac{2}{4} L^{- 1} {\frac{1}{x ^{2} + 4}}

L^{- 1} {\frac{1}{x + 2}} : e^{- 2 t}

L^{- 1} {\frac{x}{x ^{2} + 4}} : cos (2 t)

L^{- 1} {\frac{1}{x ^{2} + 4}} : \frac{1}{2} sin (2 t)

= - \frac{1}{4} e^{- 2 t} + \frac{1}{4} cos (2 t) + \frac{2}{4} \cdot \frac{1}{2} sin (2 t)

Refine

- \frac{1}{4} e^{- 2 t} + \frac{1}{4} cos (2 t) + \frac{2}{4} \frac{1}{2} sin (2 t) : - \frac{1}{4} e^{- 2 t} + \frac{1}{4} cos (2 t) + \frac{sin ( 2 t )}{4}

= - \frac{1}{4} e^{- 2 t} + \frac{1}{4} cos (2 t) + \frac{sin ( 2 t )}{4}

\int \frac{ln ( x )}{x ^{5}} d x

\int_{0}^{4} x^{2} (4 - x)^{3} d x

x \to 1 lim (\frac{1}{x ( 1 - x ) ^{2}})

\frac{d}{d x} (10 (20 x - x^{\frac{3}{2}}))

x \to 1 lim (\frac{x}{x ^{2} + 7})

What is inverse oflaplace x/((x+2)(x^2+4)) ?
The answer to inverse oflaplace x/((x+2)(x^2+4)) is -1/4 e^{-2t}+1/4 cos(2t)+(sin(2t))/4