laplacetransform 8sin(4t)-5cos(4t)

Lösung

laplace transformation

8 sin (4 t) - 5 cos (4 t)

\frac{32}{s ^{2} + 16} - \frac{5 s}{s ^{2} + 16}

Schritte zur Lösung

L {8 sin (4 t) - 5 cos (4 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)}

= 8 L {sin (4 t)} - 5 L {cos (4 t)}

L {sin (4 t)} : \frac{4}{s ^{2} + 16}

L {cos (4 t)} : \frac{s}{s ^{2} + 16}

= 8 \cdot \frac{4}{s ^{2} + 16} - 5 \cdot \frac{s}{s ^{2} + 16}

Fasse

8 \frac{4}{s ^{2} + 16} - 5 \frac{s}{s ^{2} + 16}

zusammen:

\frac{32}{s ^{2} + 16} - \frac{5 s}{s ^{2} + 16}

= \frac{32}{s ^{2} + 16} - \frac{5 s}{s ^{2} + 16}

\int_{- 2}^{2} 8 1^{\frac{x}{2}} d x

t an g e n t \frac{( x + 3 )}{( x - 3 )}

9 y^{^{''}} - y = x e^{\frac{x}{3}}, y (0) = 1, y^{^{'}} (0) = 0

\frac{\partial}{\partial y} (\frac{y}{x} + 4 x)

\int (x^{2} + x^{5}) 3 x^{6} + 2 x^{3} d x