y^{''}+2y^'+2y=g(t),y(0)=8,y^'(0)=6

Lösung

y^{^{''}} + 2 y^{^{'}} + 2 y = g (t), y (0) = 8, y^{^{'}} (0) = 6

y = e^{- t} ((8 + \frac{g}{2}) cos (t) + 14 sin (t)) + \frac{g t}{2} - \frac{g}{2}

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 2 y = g (t)

Solve linear ODE:

y = e^{- t} ((8 + \frac{g}{2}) cos (t) + 14 sin (t)) + \frac{g t}{2} - \frac{g}{2}

y = e^{- t} ((8 + \frac{g}{2}) cos (t) + 14 sin (t)) + \frac{g t}{2} - \frac{g}{2}

y^{^{''}} - 4 y = 1, y^{^{'}} (0) = 2

y^{^{''}} - 16 y = 19.2 e^{4 x} + 60 e^{x}

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 (cos^{2} (x)) e^{x}

x^{^{''}} - 4 x^{^{'}} + 4 x = 4 e^{2 t}, x^{0} = - 1, x^{^{'}} 0 = - 4

y^{^{''}} - 2 y^{^{'}} + y = 14 x^{\frac{2}{3}} e^{x}