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

The answer to inverse oflaplace (2*(s^2+1))/(s^4) is 2(t+(t^3)/6)

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inverse oflaplace (2*(s^2+1))/(s^4)

inverse laplace

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

2 (t + \frac{t ^{3}}{6})

Solution steps

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

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)}

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

L^{- 1} {\frac{s ^{2} + 1}{s ^{4}}} : t + \frac{t ^{3}}{6}

= 2 (t + \frac{t ^{3}}{6})

\frac{d}{d x} (e^{\frac{x ^{2}}{2 a}})

y^{^{'}} = y^{2} - 8

x \to 0 lim (\frac{\frac{5}{1 + x} - 5}{x})

\frac{d}{d x} (sin (2 x) d x)

(x y sin (x) + 2 y cos (x)) d x + (2 x cos (x)) d y = 0

What is inverse oflaplace (2*(s^2+1))/(s^4) ?
The answer to inverse oflaplace (2*(s^2+1))/(s^4) is 2(t+(t^3)/6)