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

Lösung

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

y = - cos (t) + c_{2} sin (t)

Schritte zur Lösung

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

Solve linear ODE:

y = - cos (t) + c_{2} sin (t)

y = - cos (t) + c_{2} sin (t)

a \cdot y^{^{''}} + b \cdot y = 0

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

y^{^{''}} + 8 y^{^{'}} + 15 y = 0, y (0) = 12, y^{^{'}} (0) = - 52

3 (\frac{d ^{2} y}{d x ^{2}}) + 19 (\frac{d y}{d x}) - 14 y = 0

(d^{3} + 8) y = 0