laplacetransform sin(2t)-5t

Lösung

laplace transformation

sin (2 t) - 5 t

\frac{2}{s ^{2} + 4} - \frac{5}{s ^{2}}

Schritte zur Lösung

L {sin (2 t) - 5 t}

Verwende die lineare Eigenschaft der Laplace-Transformation:

Für Funktionen

f (t), g (t)

und Konstanten

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

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

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

L {t} : \frac{1}{s ^{2}}

= \frac{2}{s ^{2} + 4} - 5 \cdot \frac{1}{s ^{2}}

Fasse

\frac{2}{s ^{2} + 4} - 5 \frac{1}{s ^{2}}

zusammen:

\frac{2}{s ^{2} + 4} - \frac{5}{s ^{2}}

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

\int_{1}^{3} x^{5} + \frac{1}{4 x ^{5}} d x

a re a y = x, y = 0, y = \frac{1}{x ^{2}}

x \to 0 lim ((1 - 3 x)^{\frac{7}{x}})

\int \frac{t - 3}{t + 1} d t

\int x \cdot (ln (x))^{2} d x