4y^{''}+17y=0,y(0)=3,y^'(0)=5

Lösung

4 y^{^{''}} + 17 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y = 3 cos (\frac{17 t}{2}) + \frac{10}{17} sin (\frac{17 t}{2})

Schritte zur Lösung

4 y^{''} (t) + 17 y = 0

Solve linear ODE:

y = 3 cos (\frac{17 t}{2}) + \frac{10}{17} sin (\frac{17 t}{2})

y = 3 cos (\frac{17 t}{2}) + \frac{10}{17} sin (\frac{17 t}{2})

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 1

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 7

y^{^{''''}} - 4 y^{^{'''}} + 8 y^{^{''}} - 16 y^{^{'}} + 16 y = 0

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