Question

Решите уравнение
x^2+xy'=y, y(1)=0

Answer 1

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