inverselaplace 2/(s+1)+(24s)/(s^2+9)

Lösung

inverse laplace transformation

\frac{2}{s + 1} + \frac{24 s}{s ^{2} + 9}

2 e^{- t} + 24 cos (3 t)

Schritte zur Lösung

L^{- 1} {\frac{2}{s + 1} + \frac{24 s}{s ^{2} + 9}}

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{2}{s + 1}} + L^{- 1} {\frac{24 s}{s ^{2} + 9}}

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

L^{- 1} {\frac{24 s}{s ^{2} + 9}} : 24 cos (3 t)

= 2 e^{- t} + 24 cos (3 t)

f^{^{''}} (x) = 4, f^{^{'}} (2) = 10, f (2) = 15

t an g e n t f (x) = 3 x^{3}, at x = 3

\frac{\partial}{\partial x} (x^{2} e^{y} + x - 2 y)

\int \frac{x ^{2} + 2 x + 1}{x ^{3} + x} d x

\frac{d y}{d x} = e^{y - x^{2} + 1}