inverselaplace (s-5)/(s^2-6s+25)

Lösung

inverse laplace transformation

\frac{s - 5}{s ^{2} - 6 s + 25}

e^{3 t} cos (4 t) - \frac{1}{2} e^{3 t} sin (4 t)

Schritte zur Lösung

L^{- 1} {\frac{s - 5}{s ^{2} - 6 s + 25}}

Schreibe

\frac{s - 5}{s ^{2} - 6 s + 25}

um:

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

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

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

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

= e^{3 t} cos (4 t) - 2 e^{3 t} \frac{1}{4} sin (4 t)

Fasse

e^{3 t} cos (4 t) - 2 e^{3 t} \frac{1}{4} sin (4 t)

zusammen:

e^{3 t} cos (4 t) - \frac{1}{2} e^{3 t} sin (4 t)

= e^{3 t} cos (4 t) - \frac{1}{2} e^{3 t} sin (4 t)

\int 1 + 4 a^{2} x^{2} d x

\int ln (x - 2) d x

t an g e n t y = \frac{1}{x ^{3}}, (1, 1)

t an g e n t f (x) = \frac{5 x}{x - 3}, (4, 20)

d er i v a t i v e y = (x + 5)^{\frac{1}{x}}