laplacetransform sin(4t+5pi)

Lösung

laplace transformation

sin (4 t + 5 π)

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

Schritte zur Lösung

L {sin (4 t + 5 π)}

Wende Laplace-Transformationstabelle an:

L {sin (a t + b)} = \frac{s \cdot sin ( b ) + a \cdot cos ( b )}{s ^{2} + a ^{2}}

L {sin (4 t + 5 π)} = \frac{s sin ( 5 π ) + 4 cos ( 5 π )}{s ^{2} + 4 ^{2}}

= \frac{s sin ( 5 π ) + 4 cos ( 5 π )}{s ^{2} + 4 ^{2}}

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

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\frac{\partial}{\partial u} (e^{u})

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\frac{\partial}{\partial x} (\frac{x ^{2}}{x + y})

d er i v a t i v e x \cdot ln (x)