laplacetransform e^{-t}(t^2+4t+5)

Lösung

laplace transformation

e^{- t} (t^{2} + 4 t + 5)

\frac{5 s ^{2} + 14 s + 11}{( s + 1 ) ^{3}}

Schritte zur Lösung

L {e^{- t} (t^{2} + 4 t + 5)}

Wende Transformationsregel an: wenn

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

dann

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

Für

e^{- t} (t^{2} + 4 t + 5) : f (t) = (t^{2} + 4 t + 5), a = - 1

= L {(t^{2} + 4 t + 5)} (s + 1)

L {(t^{2} + 4 t + 5)} : \frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{5}{s}

= \frac{2}{( s + 1 ) ^{3}} + \frac{4}{( s + 1 ) ^{2}} + \frac{5}{( s + 1 )}

= \frac{2}{( s + 1 ) ^{3}} + \frac{4}{( s + 1 ) ^{2}} + \frac{5}{s + 1}

Vereinfache

\frac{2}{( s + 1 ) ^{3}} + \frac{4}{( s + 1 ) ^{2}} + \frac{5}{s + 1} : \frac{5 s ^{2} + 14 s + 11}{( s + 1 ) ^{3}}

= \frac{5 s ^{2} + 14 s + 11}{( s + 1 ) ^{3}}

y^{^{'}} + 4 e^{x + y} = 0

x \to 0 lim (\frac{9 x ^{2} - 2 x}{x})

\frac{d}{d x} (arccos (- x))

\int \frac{1}{t ^{2} 441 - t ^{2}} d t

\int (4 x^{5} + 1) x^{4} d x