y^{''}+3y^'+2y=u_{2}(t),y(0)=0,y^'(0)=1

Lösung

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 0, y^{^{'}} (0) = 1

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

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

(d^{2} + 2 d + 3) y = e^{2} x sin (x) + 3

(d - 2)^{3} y = e^{2} x

y^{^{''}} - 4 y^{^{'}} + 3 y = 4 e^{x}

4 y^{^{'''}} - 9 y^{^{''}} + 14 y^{^{'}} - 3 y = x e^{x}

y^{^{''}} + y^{^{'}} + 4 y = cos (3 t)