(x^2+1)sin(y)y^'=-2xcos(y)

Lösung

(x^{2} + 1) sin (y) y^{^{'}} = - 2 x cos (y)

y = arccos (e^{- c_{1}} x^{2} + e^{- c_{1}}) + 2 πn, y = - arccos (e^{- c_{1}} x^{2} + e^{- c_{1}}) + 2 πn

Schritte zur Lösung

(x^{2} + 1) sin (y) y^{'} (x) = - 2 x cos (y)

Solve separable ODE:

y = arccos (e^{- c_{1}} x^{2} + e^{- c_{1}}) + 2 πn, y = - arccos (e^{- c_{1}} x^{2} + e^{- c_{1}}) + 2 πn

y = arccos (e^{- c_{1}} x^{2} + e^{- c_{1}}) + 2 πn, y = - arccos (e^{- c_{1}} x^{2} + e^{- c_{1}}) + 2 πn

\frac{d y}{d t} = - 3 y

y^{^{'}} = \frac{x ^{7}}{y ^{8}}

\frac{d y}{d x} = k \cdot a \cdot y

\frac{d y}{d x} = e^{x + y + 5}

\frac{d y}{d x} (x^{2} + 1) = - 2 x y