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

Решите задачу:

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

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