y=u*v ; y'=u'v+uv'
u'v+uv'+uv=x
u'v + u * (v' + v) = x
1) v' + v = 0
dv/dx = -v
dv/v = -dx
int dv/v = -int dx
ln |v| = -x
v = e^(-x)
2) u' * e^(-x) = x
u' = x * e^x
u = int x * e^x dx = int x d(e^x) = x * e^x - int e^x dx = x * e^x - e^x + C
Тогда y = (x * e^x - e^x + C) * e^(-x) = x - 1 + C * e^(-x)
y(0) = 0 => 0 = -1 + C => C = 1
Ответ: y = x - 1 + e^(-x)