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

Lösung

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

y = - \frac{1}{- arctan ( x ) + c _{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{1}{- arctan ( x ) + c _{1}}

y = - \frac{1}{- arctan ( x ) + c _{1}}

x^{2} d y - y^{2} d x = 0

y^{^{'}} = \frac{( 3 t ^{2} + 4 t + 2 )}{( 2 y - 2 )}

y tan (x) \frac{d y}{d x} = (4 + y^{2}) sec (x^{2})

\frac{d y}{d x} = c \cdot y

tan (x) d y + 2 y d x = 0