What is inverse oflaplace ((x))/((x+2)(x-3)) ?

The answer to inverse oflaplace ((x))/((x+2)(x-3)) is 2/5 e^{-2t}+3/5 e^{3t}

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inverse oflaplace ((x))/((x+2)(x-3))

inverse laplace

\frac{( x )}{( x + 2 ) ( x - 3 )}

\frac{2}{5} e^{- 2 t} + \frac{3}{5} e^{3 t}

Solution steps

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

Take the partial fraction of

\frac{x}{( x + 2 ) ( x - 3 )} : \frac{2}{5 ( x + 2 )} + \frac{3}{5 ( x - 3 )}

= L^{- 1} {\frac{2}{5 ( x + 2 )} + \frac{3}{5 ( x - 3 )}}

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{2}{5 ( x + 2 )}} + L^{- 1} {\frac{3}{5 ( x - 3 )}}

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

L^{- 1} {\frac{3}{5 ( x - 3 )}} : \frac{3}{5} e^{3 t}

= \frac{2}{5} e^{- 2 t} + \frac{3}{5} e^{3 t}

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

x \frac{d y}{d x} = y^{2} - y

\int_{1}^{4} t^{- \frac{3}{2}} d t

x^{^{''}} + 10 x^{^{'}} + 25 = 14 e^{- 5 t}

\frac{d y}{d t} = - 4 (y - 3)

What is inverse oflaplace ((x))/((x+2)(x-3)) ?
The answer to inverse oflaplace ((x))/((x+2)(x-3)) is 2/5 e^{-2t}+3/5 e^{3t}