Помогите решить!x2y' = y(x + y)

[latex] x^{2} y^{'}=y(x+y) [/latex]   дифференциальное уравнение [latex]y^{'}= \frac{y}{x}+( \frac{y}{x} )^2 [/latex] замена: [latex]t(x)= \frac{y}{x};y=t(x)x;y^{'}=t^{'}x+t [/latex] [latex]t^{'}x+t=t+t^2[/latex] [latex]t^{'}x=t^2[/latex] [latex] \frac{dt}{dx} x=t^2[/latex] [latex] \frac{dt}{t^2}= \frac{dx}{x} [/latex] [latex]- \frac{1}{t} =ln!x!+C[/latex] [latex]t=- \frac{1}{ln!x!+C} [/latex] [latex] \frac{y}{x}=- \frac{1}{ln!x!+C} [/latex] [latex]y=- \frac{x}{ln!x!+C} [/latex]