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

Lösung

inverse laplace transformation

(\frac{2}{s} - \frac{1}{s ^{3}})^{2}

4 t - \frac{2 t ^{3}}{3} + \frac{t ^{5}}{120}

Schritte zur Lösung

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

Schreibe

(\frac{2}{s} - \frac{1}{s ^{3}})^{2}

um:

\frac{4}{s ^{2}} - \frac{4}{s ^{4}} + \frac{1}{s ^{6}}

= L^{- 1} {\frac{4}{s ^{2}} - \frac{4}{s ^{4}} + \frac{1}{s ^{6}}}

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

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

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

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

L^{- 1} {\frac{1}{s ^{6}}} : \frac{t ^{5}}{120}

= 4 t - \frac{2 t ^{3}}{3} + \frac{t ^{5}}{120}

a re a y = x + 3, y = \frac{( x + 3 )}{2}

(1 - x^{3}) \frac{d y}{d x} = 3 x^{2} y

\frac{\partial}{\partial x} (8 y)

\int x (x^{2} + 4) d x

\int 3 x^{2} + 5 d x