1/8 y^{''}+4y^'+12y=0,y(0)=0,y^'(0)=4

Lösung

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 4

y = \frac{1}{2 10} e^{4 (10 - 4) t} - \frac{1}{2 10} e^{- 4 (4 + 10) t}

Schritte zur Lösung

\frac{1}{8} y^{''} (t) + 4 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = \frac{1}{2 10} e^{4 (10 - 4) t} - \frac{1}{2 10} e^{- 4 (4 + 10) t}

y = \frac{1}{2 10} e^{4 (10 - 4) t} - \frac{1}{2 10} e^{- 4 (4 + 10) t}

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

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 3, y^{^{'}} (0) = - 11

y^{^{''}} - 6 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = - 1, y^{^{'}} (0) = 0

2 x^{^{''}} + 2 x^{^{'}} + 7 x = 0