xy' = (y^2) * lnx - y
1) xy' + y = 0
x * dy/dx = -y
dy/y = -dx/x
ln |y| = -ln |x|
y = C/x
2) y = C(x)/x
x * (C' * x - C)/x^2 = C^2/x^2 * ln x - C/x
C' - C/x = C^2/x^2 * ln x - C/x
C' = C^2/x^2 * ln x
dC/dx = C^2/x^2 * ln x
dC/C^2 = ln x/x^2 dx
-1/C = int ln x/x^2 dx
int ln x/x^2 dx = int ln x d(-1/x) = -ln x/x + int 1/x d(ln x) = -ln x/x + int 1/x^2 dx = -ln x/x - 1/x + C
Тогда
C(x) = 1/(ln x/x + 1/x + C)
y = 1/(ln x + 1 + Cx)
Ответ: y = 1/(ln x + 1 + Cx) и y = 0.
Кажется так, если не напутал.