inverselaplace 1/((s^2+6s+25))

Lösung

inverse laplace transformation

\frac{1}{( s ^{2} + 6 s + 25 )}

e^{- 3 t} \frac{1}{4} sin (4 t)

Schritte zur Lösung

L^{- 1} {\frac{1}{( s ^{2} + 6 s + 25 )}}

\frac{1}{s ^{2} + 6 s + 25} = \frac{1}{( s + 3 ) ^{2} + 16}

= L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 16}}

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} + 16} : a = - 3, F (s) = \frac{1}{s ^{2} + 16}

= e^{- 3 t} L^{- 1} {\frac{1}{s ^{2} + 16}}

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

= e^{- 3 t} \frac{1}{4} sin (4 t)

\int 4 sin (t) d t

y^{^{''}} + 9 y = t^{2} e^{3 t} + 6

\frac{\partial}{\partial y} (x y sin (z))

x \to 0 lim (\frac{e ^{- \frac{1}{x}}}{x})

\int_{3}^{3.5} \frac{x}{8} d x