y^{''}+3y^'+2y=u_{2}(t),y(0)=5,y^'(0)=3

Lösung

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 5, y^{^{'}} (0) = 3

y = e^{- t} (5 + \frac{u _{2} + 32}{4} + \frac{3 u _{2}}{4}) - \frac{e ^{- 2 t} ( u _{2} + 32 )}{4} + \frac{u _{2} t}{2} - \frac{3 u _{2}}{4}

Schritte zur Lösung

y^{''} (t) + 3 y^{'} (t) + 2 y = u_{2} (t)

Solve linear ODE:

y = e^{- t} (5 + \frac{u _{2} + 32}{4} + \frac{3 u _{2}}{4}) - \frac{e ^{- 2 t} ( u _{2} + 32 )}{4} + \frac{u _{2} t}{2} - \frac{3 u _{2}}{4}

y = e^{- t} (5 + \frac{u _{2} + 32}{4} + \frac{3 u _{2}}{4}) - \frac{e ^{- 2 t} ( u _{2} + 32 )}{4} + \frac{u _{2} t}{2} - \frac{3 u _{2}}{4}

(D^{2} - D) y = e^{x} sin (x)

(D^{3} + 2 D^{2} - 5 D - 6) y = e^{- 3 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = sin (2 t)

y^{^{''}} + 16 y = 17 sec^{2} (4 t)

y^{^{''}} - y = t, y^{^{'}} (0) = 1