Question

Высшая математика. СРОЧНО!!!!

Answer 1

$y'-\dfrac{y}{x}=cos^2(2\dfrac{y}{x})\\ \left[y=xz=>y'=z+xz'\right]\\ z+xz'-z=cos^2(2z)\\xz'=cos^2(2z) \\\dfrac{1}{2} \int\dfrac{2dz}{cos^2(2z)}=\int\dfrac{dx}{x}\\ \dfrac{1}{2} tg(2z)=lnCx\\ tg(2\dfrac{y}{x})=2lnCx$