inverselaplace (5s)/(9+s^2)

Lösung

inverse laplace transformation

\frac{5 s}{9 + s ^{2}}

5 cos (3 t)

Schritte zur Lösung

L^{- 1} {\frac{5 s}{9 + s ^{2}}}

Wende inverse Transformationsregel an: if

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

dann

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

Für

\frac{5 s}{9 + s ^{2}} : F (s) = \frac{5}{9 + s ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{5}{9 + s ^{2}}}) + L^{- 1} {\frac{5}{9 + s ^{2}}} (0)

L^{- 1} {\frac{5}{9 + s ^{2}}} : \frac{5}{3} sin (3 t)

\frac{d}{d t} (\frac{5}{3} sin (3 t)) = 5 cos (3 t)

Evaluiere

\frac{5}{3} sin (3 t)

am Punkt

0 = 0

= 5 cos (3 t)

d er i v a t i v e \frac{5 x ^{3} + 3 x ^{2}}{x}

\frac{\partial}{\partial x} (x^{2} (1 - y^{\frac{2}{5}}))

d er i v a t i v e y = x + x^{\frac{3}{2}} - 8 x

(5 x^{3} + 3 x y + 2 y^{2}) d x + (x^{2} + 2 x y) d y = 0

\frac{d}{d x} (ln (x^{2} + 2 x))