а) x * y' + y = 0
x * dy/dx + y = 0
x * dy/dx = -y
dy/y = -dx/x
int dy/y = - int dx/x
ln |y| = -ln |x| + C => y = C/x.
б) x * y' + y = y^2, y(x) = C(x)/x
Подставляем и получаем:
x * (C'(x) * x - C(x))/x^2 + C(x)/x = C^2(x)/x^2
C'(x) = C^2(x)/x^2
dC(x)/dx = C^2(x)/x^2
dC(x)/C^2(x) = dx/x^2
-1/C(x) = -1/x + C => 1/C(x) = 1/x + C => C(x) = x/(1 + Cx)
Тогда
y = 1/(1 + Cx)
y(1) = 1/2 => 1/2 = 1/(1 + C) => C = 1

Ответ: y = 1/(1 + x).