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

Lösung

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

y = cos (2 t) - \frac{cos ( 2 π ) sin ( 2 t )}{sin ( 2 π )}

Schritte zur Lösung

y^{''} (t) + 2 y = 0

Solve linear ODE:

y = cos (2 t) - \frac{cos ( 2 π ) sin ( 2 t )}{sin ( 2 π )}

y = cos (2 t) - \frac{cos ( 2 π ) sin ( 2 t )}{sin ( 2 π )}

3 y^{^{''}} - 15 y^{^{'}} + 4 y = 0

1 y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 27 y = 0, y (0) = 1, y^{^{'}} (0) = - 12

18 y^{^{''}} + 9 y^{^{'}} - 5 y = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = 4