laplacetransform e^{-t}(t-1)

Lösung

laplace transformation

e^{- t} (t - 1)

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

Schritte zur Lösung

L {e^{- t} (t - 1)}

Wende Transformationsregel an: wenn

L {f (t)} = F (s)

dann

L {e^{a t} f (t)} = F (s - a)

Für

e^{- t} (t - 1) : f (t) = (t - 1), a = - 1

= L {(t - 1)} (s + 1)

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

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

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

Vereinfache

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

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

d er i v a t i v e f (x) = sin^{4} (x)

\frac{d}{d x} (100 x^{2})

\frac{\partial}{\partial x} (x ln (yz))

\int x e^{a x} d x

2 y^{^{''}} - 3 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = \frac{1}{2}