inverselaplace 1/((s^2-4s+5))

Lösung

inverse laplace transformation

\frac{1}{( s ^{2} - 4 s + 5 )}

e^{2 t} sin (t)

Schritte zur Lösung

L^{- 1} {\frac{1}{( s ^{2} - 4 s + 5 )}}

\frac{1}{s ^{2} - 4 s + 5} = \frac{1}{( s - 2 ) ^{2} + 1}

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

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

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

= e^{2 t} sin (t)

x^{2} \frac{d y}{d x} = y

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

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

x^{2} + 2 y^{2} - 2 x - 8 y + 6 = 0 (0.3)

\int \frac{36 x ^{2} - 49}{x ^{3}} d x