inverselaplace 3/(5(s-3)^2+80)

Lösung

inverse laplace transformation

\frac{3}{5 ( s - 3 ) ^{2} + 80}

e^{3 t} \frac{3}{20} sin (4 t)

Schritte zur Lösung

L^{- 1} {\frac{3}{5 ( s - 3 ) ^{2} + 80}}

Wende inverse Transformationsregel an: if

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

dann

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

Für

\frac{3}{5 ( s - 3 ) ^{2} + 80} : a = 3, F (s) = \frac{3}{5 s ^{2} + 80}

= e^{3 t} L^{- 1} {\frac{3}{5 s ^{2} + 80}}

L^{- 1} {\frac{3}{5 s ^{2} + 80}} : \frac{3}{20} sin (4 t)

= e^{3 t} \frac{3}{20} sin (4 t)

a re a x^{2} + 2 x + 1, 2 x + 5, - 2, 2

\int x^{2} + 4 x + 5 d x

y d y = 4 x y^{2} + 1 d x

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x ^{2}}

(\frac{x}{x ^{2} + y ^{4}}) d x + (\frac{( 2 y ^{3} )}{x ^{2} + y ^{4}}) d y = 0