laplacetransform e^{-2t-5}

Lösung

laplace transformation

e^{- 2 t - 5}

\frac{1}{e ^{5} ( s + 2 )}

Schritte zur Lösung

L {e^{- 2 t - 5}}

Verwende die konstante Multiplikationseigenschaft der Laplace-Transformation:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= e^{- 5} L {e^{- 2 t}}

L {e^{- 2 t}} : \frac{1}{s + 2}

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

Fasse

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

zusammen:

\frac{1}{e ^{5} ( s + 2 )}

= \frac{1}{e ^{5} ( s + 2 )}

\frac{\partial}{\partial x} (y \cdot ln (x))

\int \frac{x + 1}{x ^{3} - x ^{2}} d x

\int_{0}^{π} sin^{2} (x) d x

\frac{\partial}{\partial x} (ln (4 - x^{2}))

t an g e n t y = 4 x^{2} - x^{3}, (1, 3)