Question

Дифференциальные уравнения xy'+y=y^2

Answer 1

$xy'+y=y^2\\\\xy'=y^2-y\\\\y'=\frac{y^2-y}{x}\\\\\frac{dy}{dx}=\frac{y^2-y}{x}\\\\\int \frac{dy}{y^2-y}=\int \frac{dx}{x}\\\\\int \frac{dy}{(y-\frac{1}{2})^2-\frac{1}{4}}=\int \frac{dx}{x}\\\\\frac{1}{2\cdot \frac{1}{2}}\cdot ln\left |\frac{y-\frac{1}{2}-\frac{1}{2}}{y-\frac{1}{2}+\frac{1}{2}}\right |=ln|x|+lnC\\\\ln\left |\frac{y-1}{y}\right |=ln|x|+lnC\\\\\frac{y-1}{y}=cx\\\\1-\frac{1}{y}=cx\\\\\frac{1}{y}=1-cx\\\\y=\frac{1}{1-cx}$