inverselaplace 1/((s^2-6s+10))

Lösung

inverse laplace transformation

\frac{1}{( s ^{2} - 6 s + 10 )}

e^{3 t} sin (t)

Schritte zur Lösung

L^{- 1} {\frac{1}{( s ^{2} - 6 s + 10 )}}

\frac{1}{s ^{2} - 6 s + 10} = \frac{1}{( s - 3 ) ^{2} + 1}

= L^{- 1} {\frac{1}{( s - 3 ) ^{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 - 3 ) ^{2} + 1} : a = 3, F (s) = \frac{1}{s ^{2} + 1}

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

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

= e^{3 t} sin (t)

\frac{\partial ^{2}}{\partial x ^{2}} (e^{3 x + 4 y} cos (5 z))

in v erse l a pl a ce \frac{2}{5 ( s + 3 ) ^{2} + 125}

\frac{d}{d x} (\frac{x ^{2}}{x ^{3} + 1})

f (x) = 4 sin (4 x)

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