What is inverse oflaplace 4/(2(s-1)^2+50) ?

The answer to inverse oflaplace 4/(2(s-1)^2+50) is e^t 2/5 sin(5t)

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inverse oflaplace 4/(2(s-1)^2+50)

inverse laplace

\frac{4}{2 ( s - 1 ) ^{2} + 50}

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

Solution steps

L^{- 1} {\frac{4}{2 ( s - 1 ) ^{2} + 50}}

Apply inverse transform rule:

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

then

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

For

\frac{4}{2 ( s - 1 ) ^{2} + 50} : a = 1, F (s) = \frac{4}{2 s ^{2} + 50}

= e^{t} L^{- 1} {\frac{4}{2 s ^{2} + 50}}

L^{- 1} {\frac{4}{2 s ^{2} + 50}} : \frac{2}{5} sin (5 t)

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

\frac{d}{d x} (12 sec (x))

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

4 x^{2} - 4, x^{8} - 1

\frac{\partial}{\partial x} (y + 3 x z + x^{3})

f (x) = x^{3} - 4 x^{2} - 9 x + 11

What is inverse oflaplace 4/(2(s-1)^2+50) ?
The answer to inverse oflaplace 4/(2(s-1)^2+50) is e^t 2/5 sin(5t)