y^{''}+6y^'+9y=(-16e^{-3t})/(t^2+1)

Lösung

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{- 16 e ^{- 3 t}}{t ^{2} + 1}

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + 8 e^{- 3 t} ln (t^{2} + 1) - 16 e^{- 3 t} t arctan (t)

Schritte zur Lösung

y^{''} (t) + 6 y^{'} (t) + 9 y = \frac{- 16 e ^{- 3 t}}{t ^{2} + 1}

Solve linear ODE:

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + 8 e^{- 3 t} ln (t^{2} + 1) - 16 e^{- 3 t} t arctan (t)

y = c_{1} e^{- 3 t} + c_{2} t e^{- 3 t} + 8 e^{- 3 t} ln (t^{2} + 1) - 16 e^{- 3 t} t arctan (t)

in v erse l a pl a ce \frac{s + 5}{s ^{2} + 5 s + 6}

\frac{d}{d x} (\frac{x - 1}{x ^{2} + 2 x + 1})

\int \frac{2 x}{( x ^{2} + 1 )} d x

\frac{d}{d x} (\frac{x ^{2} - 4 x + 3}{3 x})

\frac{\partial ^{2}}{\partial y ^{2}} (x y^{2} - 6 x^{2} - 3 y^{2})