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

Lösung

inverse laplace transformation

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

e^{t} sin (t)

Schritte zur Lösung

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

Wende inverse Transformationsregel an: if

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

dann

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

Für

\frac{1}{( s - 1 ) ^{2} + 1} : a = 1, F (s) = \frac{1}{s ^{2} + 1}

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

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

= e^{t} sin (t)

t an g e n t f (x) = csc (x), at x = \frac{π}{6}

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

d er i v a t i v e \frac{( x + 1 ) ( x ^{2} + 1 )}{x - 2}

\int_{- 1}^{1} (6 x^{5} + 3) d x

d er i v a t i v e y = \frac{9 x}{e ^{x}}