Question

sin^{2} x* cos^{2} ydy-cos2xdy=0 y(0)= frac{ pi }{4}

Answer 1

$sin^2xcdot cos^2y, dy-cos2x, dx=0; ,quad y(frac{pi}{4})=0\\sin^2xcdot cos^2y, dy=cos2x, dx\\int cos^2y, dy =int frac{cos2x, dx}{sin^2x} \\int frac{1+cos2y}{2}dy=int frac{1-2sin^2x}{sin^2x}dx\\frac{1}{2}cdot int (1+cos2y)dy=int ( frac{1}{sin^2x} -2)dx\\frac{1}{2}cdot (y+frac{1}{2}sin2y)=-ctgx-2x+C\\y(frac{pi}{4})=0; ,quad frac{1}{2}cdot (0+frac{1}{2}cdot 1)=-1-frac{pi}{2}+C\\C=frac{5+2pi }{4}$

$frac{1}{2}cdot (y+frac{1}{2}sin2y)=-ctgx-2x+frac{5+2pi}{4}$