Question

$y'+2xy=xe^{-x^{2} }$ Решить методом Бернулли

Answer 1

y'+2xy=x e^(-x^2)

[y=uv=>y'=u'v+uv']

u'v+uv'+2xuv=x e^(-x^2)

u'v+u(v'+2xv)=x e^(-x^2)

(*) v'+2xv=0 => dv/v=-2xdx => v=e^(-x^2)

u' e^(-x^2)=x e^(-x^2)

u'=x

u=(x^2)/2+C

y=((x^2)/2+C) e^(-x^2)