laplacetransform e^{2t}tsin(5t)

Lösung

laplace transformation

e^{2 t} t sin (5 t)

\frac{5 ( 2 s - 4 )}{( s ^{2} - 4 s + 29 ) ^{2}}

Schritte zur Lösung

L {e^{2 t} t sin (5 t)}

Wende Laplace-Transformationstabelle an:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

Für

e^{2 t} t sin (5 t) : f (t) = e^{2 t} sin (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{2 t} sin (5 t)})

L {e^{2 t} sin (5 t)} : \frac{5}{( s - 2 ) ^{2} + 25}

\frac{d}{d s} (\frac{5}{( s - 2 ) ^{2} + 25}) = - \frac{5 ( 2 s - 4 )}{( s ^{2} - 4 s + 29 ) ^{2}}

= (- 1)^{1} (- \frac{5 ( 2 s - 4 )}{( s ^{2} - 4 s + 29 ) ^{2}})

= \frac{5 ( 2 s - 4 )}{( s ^{2} - 4 s + 29 ) ^{2}}

n = 0 \sum \infty (- 1)^{n} x^{2 n}

t an g e n t y = \frac{3 x + 1}{x ^{2} + 1}, (1, 2)

a re a y = x^{2} - 3 x, y = 5 x

y^{^{'}} = 5 y + y^{5}, y (0) = 1

x \to 2 - lim (x^{2} + 3 x)