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

Lösung

laplace transformation

e^{2 t} sin (5 t)

\frac{5}{( s - 2 ) ^{2} + 25}

Schritte zur Lösung

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

Wende Transformationsregel an: wenn

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

dann

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

Für

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

= L {sin (5 t)} (s - 2)

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

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

\int_{0}^{\frac{π}{2}} - 2 sin^{2} (3 x) d x

(\frac{d y}{d x}) = 0.011 \cdot y \cdot (2 - (y - 25))

\int \frac{5}{( 3 x - 1 ) ^{7}} d x

\frac{d ^{2}}{d x ^{2}} (4 x cos (x^{2}))

in v erse l a pl a ce \frac{4 s}{s ^{2} + 6 s + 9}