Question

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

Answer 1

xy'=y+x^2 cos(x)

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

xu'v+xuv'=uv+x^2 cos(x)

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

(*) xv'-v=0=> dv/v=dx/x => v=x

xu'x=x^2 cos(x)

u'=cos(x)

u=sin(x)+C

y=x(sin(x)+C)