ay^{''}+by^'-cy=0

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 3, y (0) = 5

y^{^{''}} + 1 y = 0, y (\frac{π}{2}) = 2, y^{^{'}} (\frac{π}{2}) = 3

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{5}{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 15, y (1) = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 28