inverselaplace (25)/(x(x^2+6x+25))

Lösung

inverse laplace transformation

\frac{25}{x ( x ^{2} + 6 x + 25 )}

H (t) - e^{- 3 t} cos (4 t) - \frac{3}{4} e^{- 3 t} sin (4 t)

Schritte zur Lösung

L^{- 1} {\frac{25}{x ( x ^{2} + 6 x + 25 )}}

Ermittle den Partialbruch von

\frac{25}{x ( x ^{2} + 6 x + 25 )} : \frac{1}{x} + \frac{- x - 6}{x ^{2} + 6 x + 25}

= L^{- 1} {\frac{1}{x} + \frac{- x - 6}{x ^{2} + 6 x + 25}}

Schreibe

\frac{- x - 6}{x ^{2} + 6 x + 25}

um:

- \frac{x + 3}{( x + 3 ) ^{2} + 16} - 3 \cdot \frac{1}{( x + 3 ) ^{2} + 16}

= L^{- 1} {\frac{1}{x} - \frac{x + 3}{( x + 3 ) ^{2} + 16} - 3 \cdot \frac{1}{( x + 3 ) ^{2} + 16}}

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{x}} - L^{- 1} {\frac{x + 3}{( x + 3 ) ^{2} + 16}} - 3 L^{- 1} {\frac{1}{( x + 3 ) ^{2} + 16}}

L^{- 1} {\frac{1}{x}} : H (t)

L^{- 1} {\frac{x + 3}{( x + 3 ) ^{2} + 16}} : e^{- 3 t} cos (4 t)

L^{- 1} {\frac{1}{( x + 3 ) ^{2} + 16}} : e^{- 3 t} \frac{1}{4} sin (4 t)

= H (t) - e^{- 3 t} cos (4 t) - 3 e^{- 3 t} \frac{1}{4} sin (4 t)

Fasse

H (t) - e^{- 3 t} cos (4 t) - 3 e^{- 3 t} \frac{1}{4} sin (4 t)

zusammen:

H (t) - e^{- 3 t} cos (4 t) - \frac{3}{4} e^{- 3 t} sin (4 t)

= H (t) - e^{- 3 t} cos (4 t) - \frac{3}{4} e^{- 3 t} sin (4 t)

\frac{d x}{d t} = x - 12

\frac{d}{d x} (sin (\frac{1}{x}) \cdot x^{3})

d er i v a t i v e cos (sin (tan (5 x)))

x \to - 1 lim (x^{3} - 1)

\int \frac{cos ( \frac{π}{x ^{15}} )}{x ^{16}} d x