Найти частные решения дифференциальных уравнений 3 вариант

Решите задачу:

$\frac{dy}{x^2}-\frac{dx}{y^2}=0|*x^2y^2\\y^2dy-x^2dx=0\\y^2dy=x^2dx\\\int y^2dy=\int x^2dx\\\frac{y^3}{3}=\frac{x^3}{3}+c\\c=\frac{8}{3}\\\frac{y^3}{3}=\frac{x^3}{3}+\frac{8}{3}\\y^3=x^3+8\\y=\sqrt[3]{x^3+8}$

$x^2y`+2xy=-4|:x^2\\y`+\frac{2y}{x}=-\frac{4}{x^2};y=uv;y`=u`v+v`u\\u`v+v`u+\frac{2uv}{x}=-\frac{4}{x^2}\\u`v+u(v`+\frac{2v}{x})=-\frac{4}{x^2}\\\begin{cases}v`+\frac{2v}{x}=0\\u`v=-\frac{4}{x^2}\end{cases}\\1)v`+\frac{2v}{x}=0\\\frac{dv}{dx}+\frac{2v}{x}=0\\\frac{dv}{dx}=-\frac{2v}{x}|*\frac{dx}{2v}\\\frac{dv}{2v}=-\frac{dx}{x}\\\frac{1}{2}\int\frac{dv}{v}=-\int\frac{dx}{x}\\\frac{ln|v|}{2}=-ln|x|\\ln|v|=ln|x|^{-2}\\v=x^{-2}\\\\2)u`x^{-2}=-\frac{4}{x^2}\\\frac{u`}{x^2}=-\frac{4}{x^2}\\u`=-4\\u=-4x+C$
$y=uv=(-4x+C)x^{-2}=\frac{-4}{x}+\frac{C}{x^2}\\y(-1)=0;0=4+C\\C=-4\\y=-\frac{4}{x}-\frac{4}{x^2}$