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

Lösung

inverse laplace transformation

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

e^{- 3 t} \cdot 9 t

Schritte zur Lösung

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

Wende inverse Transformationsregel an: if

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

dann

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

Für

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

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

L^{- 1} {\frac{9}{s ^{2}}} : 9 t

= e^{- 3 t} \cdot 9 t

\frac{d}{d x} (- \frac{4}{5} e^{(\frac{2}{3} x)})

\int \frac{2 x + 7}{( 2 x ^{2} + 14 x ) ^{4}} d x

\int sin (x) (1 + cos (x))^{4} d x

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

l a pl a ce t r an s f or m e^{- t} (\frac{e ^{t} + e ^{- t}}{2})