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

Lösung

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

x = c_{1} e^{\frac{t ( - a + a ^{2} - 16 )}{2}} + c_{2} e^{\frac{t ( - a - a ^{2} - 16 )}{2}}

Schritte zur Lösung

x^{''} (t) + a x^{'} (t) + 2^{2} x = 0

Solve linear ODE:

x = c_{1} e^{\frac{t ( - a + a ^{2} - 16 )}{2}} + c_{2} e^{\frac{t ( - a - a ^{2} - 16 )}{2}}

x = c_{1} e^{\frac{t ( - a + a ^{2} - 16 )}{2}} + c_{2} e^{\frac{t ( - a - a ^{2} - 16 )}{2}}

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

\frac{d ^{2} x}{d t ^{2}} + a \cdot \frac{d x}{d t} = 0

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 4 y = 0

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 8

(d^{2} - 2) y = 0