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

Lösung

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

y = \frac{c _{1} x - x - 1}{x + 1}, y = - \frac{c _{1} x - x - 1}{x + 1}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{c _{1} x - x - 1}{x + 1}, y = - \frac{c _{1} x - x - 1}{x + 1}

y = \frac{c _{1} x - x - 1}{x + 1}, y = - \frac{c _{1} x - x - 1}{x + 1}

v^{^{'}} = \frac{- 1}{2 ( v - 10 )}

y^{^{'}} + 4 y = 1

8 y^{^{'}} + 16 y = 0

x \cdot y + y^{^{'}} = 0

tan (y) y^{^{'}} = x (x^{2} - 1)^{- 1}