inverselaplace s/((s^2+4)^2)

Lösung

inverse laplace transformation

\frac{s}{( s ^{2} + 4 ) ^{2}}

\frac{t sin ( 2 t )}{4}

Schritte zur Lösung

L^{- 1} {\frac{s}{( s ^{2} + 4 ) ^{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{s}{( s ^{2} + 4 ) ^{2}} : F (s) = \frac{1}{( s ^{2} + 4 ) ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{1}{( s ^{2} + 4 ) ^{2}}}) + L^{- 1} {\frac{1}{( s ^{2} + 4 ) ^{2}}} (0)

L^{- 1} {\frac{1}{( s ^{2} + 4 ) ^{2}}} : \frac{1}{16} (sin (2 t) - 2 t cos (2 t))

\frac{d}{d t} (\frac{1}{16} (sin (2 t) - 2 t cos (2 t))) = \frac{t sin ( 2 t )}{4}

Evaluiere

\frac{1}{16} (sin (2 t) - 2 t cos (2 t))

am Punkt

0 = 0

= \frac{t sin ( 2 t )}{4}

t an g e n t f (x) = 6 sin (x) + 4 cos (x), at x = 0

\int \frac{( arctan ( x ) ) ^{7}}{1 + x ^{2}} d x

x \to 1 lim (\frac{1 - x}{x - 1})

d er i v a t i v e f (x) = 5 x^{4} e^{x} + e^{x} x^{5}

t a y l or 4 + x^{2} + 4 x^{4}, - 3