y^'+2x(x+1)y=0

Lösung

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

y = e^{- \frac{2 x ^{3}}{3} - x^{2} + c_{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{- \frac{2 x ^{3}}{3} - x^{2} + c_{1}}

y = e^{- \frac{2 x ^{3}}{3} - x^{2} + c_{1}}

\frac{d y}{d x} = 6 + 3 y

y y^{^{'}} = 3 cos (2 x)

y d y + 2 x d x - 1 d x = 0

\frac{d y}{d x} = y \cdot x^{2} - 1.1 y

\frac{4 t x d x}{d t} = x^{2} + 1