inverselaplace 1/(s^2+12s+100)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} + 12 s + 100}

e^{- 6 t} \frac{1}{8} sin (8 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} + 12 s + 100}}

\frac{1}{s ^{2} + 12 s + 100} = \frac{1}{( s + 6 ) ^{2} + 64}

= L^{- 1} {\frac{1}{( s + 6 ) ^{2} + 64}}

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 + 6 ) ^{2} + 64} : a = - 6, F (s) = \frac{1}{s ^{2} + 64}

= e^{- 6 t} L^{- 1} {\frac{1}{s ^{2} + 64}}

L^{- 1} {\frac{1}{s ^{2} + 64}} : \frac{1}{8} sin (8 t)

= e^{- 6 t} \frac{1}{8} sin (8 t)

\int_{0}^{1} 1 + 12 x d x

x \to 7 lim (3)

\int (4 x + 2) e^{2 x} d x

x \to 0 lim (ln (6 x + 1))

\int \frac{x ^{3} + 16}{x ^{2} + 16} d x