Решите уравнение x^2+xy'=y, y(1)=0

Решите задачу:

$x^2+xy'=y\\xy'-y=-x^2\\y=uv;y'=u'v+v'u\\xu'v+xv'u-uv=-x^2\\xu'v+u(xv'-v)=-x^2\\\begin{cases}xv'-v=0\\u'v=-x\end{cases}\\\frac{xdu}{dx}-v=0|*\frac{dx}{xv}\\\frac{dv}{v}-\frac{dx}{x}=0\\\frac{dv}{v}=\frac{dx}{x}\\\int\frac{dv}{v}=\int\frac{dx}{x}\\ln|v|=ln|x|\\v=x\\\frac{xdu}{dx}=-x|*\frac{dx}{x}\\du=-dx\\\int du=-\int dx\\u=-x+C\\y=x(-x+C)$