y^{''}-y^'+cy=0

Lösung

y^{^{''}} - y^{^{'}} + cy = 0

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

Schritte zur Lösung

y^{''} (t) - y^{'} (t) + cy = 0

Solve linear ODE:

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

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

4 y^{^{''}} - 28 y^{^{'}} + 49 y = 0

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

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

\frac{d ^{2} f}{d x ^{2}} = - a^{2} f (x)

(D^{6} + 6 D^{4} + 9 D^{2}) y = 0