laplacetransform 4t^2-5sin(3t)

Lösung

laplace transformation

4 t^{2} - 5 sin (3 t)

\frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

Schritte zur Lösung

L {4 t^{2} - 5 sin (3 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)}

= 4 L {t^{2}} - 5 L {sin (3 t)}

L {t^{2}} : \frac{2}{s ^{3}}

L {sin (3 t)} : \frac{3}{s ^{2} + 9}

= 4 \cdot \frac{2}{s ^{3}} - 5 \cdot \frac{3}{s ^{2} + 9}

Fasse

4 \frac{2}{s ^{3}} - 5 \frac{3}{s ^{2} + 9}

zusammen:

\frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

= \frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

\int_{0}^{5} \frac{3 x - 4}{x + 1} d x

x \to 0 - lim (\frac{1}{x ^{2}} - \frac{1}{x})

\int_{1}^{3} e^{x^{2}} d x

\int_{1}^{b} \frac{1}{( x + 1 ) ^{2}} d x

\int 2 e^{- 4 x} d x