inverselaplace 1/(s^2+9s+9)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} + 9 s + 9}

e^{- \frac{9 t}{2}} \frac{2}{3 5} sinh (\frac{3 5 t}{2})

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} + 9 s + 9}}

\frac{1}{s ^{2} + 9 s + 9} = \frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}

= L^{- 1} {\frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}}

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 + \frac{9}{2} ) ^{2} - \frac{45}{4}} : a = - \frac{9}{2}, F (s) = \frac{1}{s ^{2} - \frac{45}{4}}

= e^{- \frac{9}{2} t} L^{- 1} {\frac{1}{s ^{2} - \frac{45}{4}}}

L^{- 1} {\frac{1}{s ^{2} - \frac{45}{4}}} : \frac{2}{3 5} sinh (\frac{3 5 t}{2})

= e^{- \frac{9 t}{2}} \frac{2}{3 5} sinh (\frac{3 5 t}{2})

d er i v a t i v e f (t) = 2 t^{2} + 3 t - 6

\frac{\partial}{\partial x} (\frac{( a x + b y )}{c x + d y})

x y^{^{''}} + y^{^{'}} = 1

\frac{\partial}{\partial x} (6 x sin (x + 2 y - z))

\frac{\partial}{\partial y} (2 x^{2} + 2 y - 8)