y^{''}+4y^'+6y=1+e^{-t},y(0)=0,y^'(0)=0

Lösung

y^{^{''}} + 4 y^{^{'}} + 6 y = 1 + e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y = e^{- 2 t} (- \frac{1}{2} cos (2 t) - \frac{2}{3} sin (2 t)) + \frac{1}{6} + \frac{1}{3} e^{- t}

Schritte zur Lösung

y^{''} (t) + 4 y^{'} (t) + 6 y = 1 + e^{- t}

Solve linear ODE:

y = e^{- 2 t} (- \frac{1}{2} cos (2 t) - \frac{2}{3} sin (2 t)) + \frac{1}{6} + \frac{1}{3} e^{- t}

y = e^{- 2 t} (- \frac{1}{2} cos (2 t) - \frac{2}{3} sin (2 t)) + \frac{1}{6} + \frac{1}{3} e^{- t}

\frac{\partial}{\partial x} ((4 x + 7 y)^{10})

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

\int \frac{13 t ^{7} + t ^{5}}{t ^{\frac{3}{2}}} d t

x \to 0 lim (x^{x^{2}})

y^{^{'}} + 0.15 y = 1