What is inverse oflaplace 5/(2(2x+3)^2) ?

The answer to inverse oflaplace 5/(2(2x+3)^2) is (5e^{-(3t)/2}t)/(8)

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inverse oflaplace 5/(2(2x+3)^2)

inverse laplace

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

\frac{5 e ^{- \frac{3 t}{2}} t}{8}

Solution steps

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

Apply inverse transform rule:

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

then

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

For

\frac{5}{2 ( 2 x + 3 ) ^{2}} : a = - \frac{3}{2}, F (x) = \frac{5}{2 ( 2 x ) ^{2}}

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

L^{- 1} {\frac{5}{2 ( 2 x ) ^{2}}} : \frac{5 t}{8}

= \frac{5 e ^{- \frac{3 t}{2}} t}{8}

e^{x} sin (10 x)

\frac{d}{d x} (\frac{- x ^{2} + 9 x - 16}{x})

\int \frac{9 - w ^{2}}{w ^{2}} d w

\frac{d}{d y} (0)

\frac{d}{d x} (tan (x) - e^{- x})

What is inverse oflaplace 5/(2(2x+3)^2) ?
The answer to inverse oflaplace 5/(2(2x+3)^2) is (5e^{-(3t)/2}t)/(8)