Y'+y/x=3 Помогите решить

Решите задачу:

$y'+ \frac{y}{x} =3 \\ y=uv \\ y'=u'v+uv' \\ u'v+uv' +\frac{uv}{x}=3 \\ v(u'+\frac{u}{ x})+uv'=3 \\ \left \{ {{u'+\frac{u}{x}=0} \atop {uv'=3}} \right. \\ u'+\frac{u}{x}=0 \\ u'=-\frac{u}{x} \\ \frac{du}{u} =-\frac{dx}{x} \\lnu=-lnx \\ u= \frac{1}{x} \\ \frac{v'}{x}=3 \\ dv}=3xdx \\ v= \frac{3x^2}{2} +C \\ y=\frac{1}{x}(\frac{3x^2}{2} +C )$