решите уравнение:y′-y/x=xe^x, y(1)=e

Решите задачу:

$y' + \frac{y}{x} = x*e^x, \ y(1) = e\\\\ y = uv, \ y' = u'v + uv' \\\\ u'v + uv' + \frac{uv}{x} = xe^x\\\\ u'v + (v' + \frac{v}{x})u = xe^x\\\\ v' + \frac{v}{x} = 0, \ u'v = xe^x\\\\ v' = -\frac{v}{x}, \ \frac{v'}{v} = -\frac{1}{x}, \ \frac{dv}{v} = -\frac{dx}{x}\\\\ ln(v) = -ln(x), \ v = \frac{1}{x}\\\\ u'\frac{1}{x} = xe^x, \ du = x^2e^xdx, \ u = (x^2 - 2x + 2)e^x + C\\\\ y = uv = ((x^2 - 2x + 2)e^x + C)*\frac{1}{x} = \boxed{(x - 2 + \frac{2}{x})e^x + \frac{C}{x}}$

$y(1) = (1 - 2 + \frac{2}{1})e^1 + \frac{C}{1} = e + C = e \ \Rightarrow \ C = 0\\\\ \boxed{y = (x - 2 + \frac{2}{x})e^x}$