y^'=xy^2+x

Lösung

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

y = tan (\frac{x ^{2}}{2} + c_{1})

Schritte zur Lösung

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

Solve separable ODE:

y = tan (\frac{x ^{2}}{2} + c_{1})

y = tan (\frac{x ^{2}}{2} + c_{1})

\frac{d y}{d x} = - 2 y + 5

x^{3} \frac{d y}{d x} = y^{2} (x - 3)

(t^{3} + 7 t^{3} x^{2}) d t + x e^{t^{4}} d x = 0, x (0) = 0

y^{^{'}} = 3 (y - 1)^{\frac{2}{3}}

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