Question

Решить дифф. уравнение.
xy'-3y=4x^3

Answer 1

$xy'-3y=4x^3 \\ y'-\frac{3y}{x}=4x^2 \\ y=uv;y'=u'v+uv' \\ u'v+uv'-\frac{3uv}{x}=4x^2 \\ \left \{ {{v'-\frac{3v}{x}=0} \atop {u'v=4x^2}} \right. \\ \frac{dv}{dx}=\frac{3v}{x} \\ \frac{dv}{v}=3\frac{dx}{x} \\ lnv=3lnx \\ v=x^3 \\ \frac{du}{dx}x^3=4x^2 \\ du=4\frac{dx}{x} \\ u=4lnx +C\\ y=uv=x^3(4lnx+C)$