y^{''}+12y^'-25/4 y=0,y(0)=1,y^'(0)=1

Lösung

y^{^{''}} + 12 y^{^{'}} - \frac{25}{4} y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

Schritte zur Lösung

y^{''} (t) + 12 y^{'} (t) - \frac{25}{4} y = 0

Solve linear ODE:

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

y = (1 + \frac{1}{26}) e^{\frac{t}{2}} - \frac{1}{26} e^{- \frac{25 t}{2}}

x^{^{''}} + 9 x = 0, x (0) = 5, x^{^{'}} (0) = - 11

y^{^{''''}} + 5 y^{^{'''}} - 2 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

(2 d^{5} - 7 d^{4} + 12 d^{3} + 8 d^{2}) y = 0

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

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = \frac{4}{3}, y^{^{'}} (0) = \frac{1}{3}