xy^'=y^2ln(x)

Lösung

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

y = - \frac{2}{ln ^{2} ( x ) + c _{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{2}{ln ^{2} ( x ) + c _{1}}

y = - \frac{2}{ln ^{2} ( x ) + c _{1}}

(x^{3} y^{3}) d x - 3 x y^{2} d y = 0

e^{x^{2}} \frac{d y}{d x} + 2 e^{x^{2}} x y = e^{x^{2}} x

d y = \frac{8 y}{x} d x

y^{^{'}} = \frac{y - 1}{x + 1}

y d x - x d y = 2 d y, y (1) = 1