y^'+(x+1)y^3=0

Lösung

y^{^{'}} + (x + 1) y^{3} = 0

y = - \frac{1}{- x ^{2} - 2 x + c _{1}}, y = - - \frac{1}{- x ^{2} - 2 x + c _{1}}

Schritte zur Lösung

y^{'} (x) + (x + 1) y^{3} = 0

Solve separable ODE:

y = - \frac{1}{- x ^{2} - 2 x + c _{1}}, y = - - \frac{1}{- x ^{2} - 2 x + c _{1}}

y = - \frac{1}{- x ^{2} - 2 x + c _{1}}, y = - - \frac{1}{- x ^{2} - 2 x + c _{1}}

y^{^{'}} + x y^{5} = 0

y^{^{'}} = \frac{- y}{3 x}

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

y^{^{'}} + y^{5} = 0

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