y^{''}+4y^'+(a^2-2)y=0

Lösung

y^{^{''}} + 4 y^{^{'}} + (a^{2} - 2) y = 0

y = c_{1} e^{(- 2 + - a^{2} + 6) t} + c_{2} e^{(- - a^{2} + 6 - 2) t}

Schritte zur Lösung

y^{''} (t) + 4 y^{'} (t) + (a^{2} - 2) y = 0

Solve linear ODE:

y = c_{1} e^{(- 2 + - a^{2} + 6) t} + c_{2} e^{(- - a^{2} + 6 - 2) t}

y = c_{1} e^{(- 2 + - a^{2} + 6) t} + c_{2} e^{(- - a^{2} + 6 - 2) t}

l in e a r y^{^{''''}} - y = 0

25 x^{^{''}} + 5 x^{^{'}} + 204.167 x = 0, x (0) = 28.577

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 4

\frac{d ^{2} y}{d t ^{2}} - 14 \frac{d y}{d t} + 149 y = 0

4 y^{^{''}} - y^{^{'}} - y = 0