y^{''}-7y^'+15y=0

Lösung

y^{^{''}} - 7 y^{^{'}} + 15 y = 0

y = e^{\frac{7 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2}))

Schritte zur Lösung

y^{''} (t) - 7 y^{'} (t) + 15 y = 0

Solve linear ODE:

y = e^{\frac{7 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2}))

y = e^{\frac{7 t}{2}} (c_{1} cos (\frac{11 t}{2}) + c_{2} sin (\frac{11 t}{2}))

y^{^{''}} + 4 y^{^{'}} + 2.5 y = 0

(d^{3} + d^{2} + 4 d + 4) \cdot y = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

(d^{3} - 5 d^{2} + 8 d - 4) \cdot y = 0

(d^{3} - d^{2} - d + 1) y = 0