inverselaplace 3/((s+2)^2+9)

Lösung

inverse laplace transformation

\frac{3}{( s + 2 ) ^{2} + 9}

e^{- 2 t} sin (3 t)

Schritte zur Lösung

L^{- 1} {\frac{3}{( s + 2 ) ^{2} + 9}}

Wende inverse Transformationsregel an: if

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

dann

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

Für

\frac{3}{( s + 2 ) ^{2} + 9} : a = - 2, F (s) = \frac{3}{s ^{2} + 9}

= e^{- 2 t} L^{- 1} {\frac{3}{s ^{2} + 9}}

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

= e^{- 2 t} sin (3 t)

\int_{0}^{π} sin^{5} (\frac{x}{2}) d x

t \to π lim (sin (\frac{t}{6}))

\frac{d y}{d x} = 2 y 16 - x^{2}, y (0) = 1

\frac{d}{d x} (ln (16 - x^{2}))

\int 2 (2 + x^{2})^{\frac{3}{2}} d x