inverselaplace (2s+1)/(s^3)

الحلّ

inverse laplace

\frac{2 s + 1}{s ^{3}}

2 t + \frac{t ^{2}}{2}

خطوات الحلّ

L^{- 1} {\frac{2 s + 1}{s ^{3}}}

وسّع

= L^{- 1} {\frac{2 s}{s ^{3}} + \frac{1}{s ^{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)}

= 2 L^{- 1} {\frac{s}{s ^{3}}} + L^{- 1} {\frac{1}{s ^{3}}}

L^{- 1} {\frac{s}{s ^{3}}} : t

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

= 2 t + \frac{t ^{2}}{2}

h (x) = \frac{x ^{2} + 4 x + 1}{3 x + 5}

\frac{d}{d x} (5 x^{2 x})

\int 2 x \cdot e^{2 x} d x

\int (1 + \frac{1}{u})^{- 3} (\frac{1}{u ^{2}}) d u

\frac{d}{d x} (4 x + 8 x^{\frac{3}{8}})