Problem:$$ sin(2x)dx+cos(6y)dy = 0 $$ Initial value: $$ y(\frac{π}{2})=\frac{π}{6} $$
Integrate both sides: $$ \frac16sin(6x)= \frac12cos(2x)+c $$ $$ sin(6y)= 3cos(2x)+c $$ Use initial condition to find: c=3 $$ sin(6y)=3cos(2x)+3 $$ Use trig IDs $$ sin(6y)=6cos^2(x) $$ Solve for y $$ y=\frac16sin^{-1}(6cos^2(x)) $$ Is this a valid solution?
Given $$\sin2x\ dx+\cos 6y\ dy=0,\ \ y\left(\dfrac{\pi}{2}\right)=\dfrac{\pi}{6}$$
$$\cos6y\dfrac{dy}{dx}=-\sin2x$$ $$\dfrac{d}{dx}\left(\dfrac16\sin6y\right)=\dfrac{d}{dx}(\cos^2x)$$Integrate both sides with respect to $x$ $$\dfrac16\sin6y=\cos^2x+C$$$$\sin6y=6\cos^2x+C$$ We just obtained an implicit formula for the general solution. To determine $C$ set $x=\dfrac{\pi}{2}$ and $y=\dfrac{\pi}{6}$$$\sin\pi=6\cos^2\left(\dfrac{\pi}{2}\right)+C$$$$0=0+C$$$$C+0$$ Hence the solution of the initial value problem is given implicitly by $$\sin6y=6\cos^2x$$To solve for y we must exercise a little caution. The answer $y(x)=\dfrac16\arcsin(6\cos^2x)$ is wrong because then $y(\pi/2)=0$, not $\pi/6$
To match $y(\pi/2)=\pi/6$ we must choose $y=\dfrac{\pi}{6}-\dfrac16\arcsin(6\cos^2x)$.