Find the solution of y'sin 2x=2(y+cos x) that remains…

Find the solution of $y'\sin 2x=2(y+\cos x)$ that remains bounded as $x\to\dfrac{\pi}{2}$.

Step 1: Reduce to standard linear form

Dividing by $2\sin x\cos x$: $y'-\dfrac{y}{\sin x\cos x}=\dfrac{1}{\sin x}$, i.e. $y'-\dfrac{2y}{\sin 2x}=\dfrac{1}{\sin x}$.

Step 2: Find the integrating factor

$\mu(x)=e^{-\int\frac{2}{\sin2x}dx}=\csc2x+\cot2x=\dfrac{1+\cos2x}{\sin2x}=\cot x$ (using $1+\cos2x=2\cos^2x$, $\sin2x=2\sin x\cos x$).

Step 3: Multiply and integrate

$\dfrac{d}{dx}(y\cot x)=\dfrac{\cos x}{\sin^2x}$. With $u=\sin x$: $\displaystyle\int\frac{\cos x}{\sin^2x}dx=-\csc x+C$, so $y\cot x=-\csc x+C$, i.e. $y=C\tan x-\sec x$.

Step 4: Select the bounded branch

As $x\to\pi/2$, both $\tan x$ and $\sec x\to\infty$; they cancel only if $C=1$. Then $y=\tan x-\sec x=\dfrac{\sin x-1}{\cos x}$, and by L'Hôpital $\displaystyle\lim_{x\to\pi/2}\frac{\sin x-1}{\cos x}=0$, so this branch is bounded.