Решите уравнение, пожалуйста: x * dy/dx - x^2 + 2y = 0

Решите задачу:

$x\cdot \frac{dy}{dx}-x^2+2y=0\; |:x\\\\\frac{dy}{dx}-x+\frac{2y}{x}=0\\\\y'+\frac{2}{x}\cdot y=x\\\\y=uv\; ,\; y'=u'v+uv'\\\\u'v+(uv'+\frac{2}{x}\cdot uv)=x\\\\u'v+u(v'+\frac{2}{x}\cdot v)=x\\\\a)\; \; v'+\frac{2}{x}\cdot v=0\\\\\frac{dv}{dx}=-\frac{2}{x}\cdot v\\\\\int \frac{dv}{v}=-2\cdot \int \frac{dx}{x}\\\\lnv=-2\cdot lnx\; \; \to \; \; \; lnv=lnx^{-2}\; ,\; \; \; v=x^{-2}\\\\b)\; \; u'v=x\; \; \to \; \; \; u'\cdot x^{-2}=x\\\\\frac{du}{dx}=\frac{x}{x^{-2}}\; \; \to \; \; \; \int du=\int x^3\; dx$

$u=\frac{x^4}{4}+C\\\\c)\; \; y=x^{-2}\cdot (\frac{x^4}{4}+C)=\frac{1}{x^2}\cdot ( \frac{x^4}{4}+C) \\\\y=\frac{x^2}{4}+\frac{C}{x^2}$