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

Lösung

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

y = e^{- t} (3 cos (2 t) + 4 sin (2 t))

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- t} (3 cos (2 t) + 4 sin (2 t))

y = e^{- t} (3 cos (2 t) + 4 sin (2 t))

y^{^{''}} + n^{2} y = 0

l in e a r \frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 10 y = 0

θ^{^{''}} (t) + \frac{4 k}{3} θ (t) = 0

y^{^{''}} + 144 y = 0, y (0) = 3, y^{^{'}} (0) = - 48

y^{^{''}} + 4 y^{^{'}} - 12 y = 0, y (0) = 7, y^{^{'}} (0) = - 7