inverselaplace 1/(sqrt(2s+3))

Lösung

inverse laplace transformation

\frac{1}{2 s + 3}

\frac{e ^{- \frac{3 t}{2}}}{2 π t}

Schritte zur Lösung

L^{- 1} {\frac{1}{2 s + 3}}

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {F (s - a)} = e^{a t} f (t)

Für

\frac{1}{2 s + 3} : a = - \frac{3}{2}, F (s) = \frac{1}{2 s}

= e^{- \frac{3}{2} t} L^{- 1} {\frac{1}{2 s}}

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

= \frac{e ^{- \frac{3 t}{2}}}{2 π t}

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

x \to \infty lim (\frac{sin ( 13 x )}{3 x})

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

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\int \frac{5}{( 5 x + 2 ) ln ( 5 x + 2 )} d x