By using a suitable substitution, bring the equation…
By using a suitable substitution, bring the equation $x^2y'-\cos2x=1$ to a form with separated variables, and then integrate it, given $y(\infty)=\dfrac{9\pi}{4}$.
Explanation of Method
The equation is already directly integrable once solved for $y'$. Using $1+\cos2x=2\cos^2x$ does not remove the need for integration by parts; the resulting antiderivative involves the sine-integral function $\operatorname{Si}(t)=\displaystyle\int_0^t\frac{\sin u}{u}\,du$, whose known limit $\operatorname{Si}(t)\to\pi/2$ as $t\to\infty$ lets the condition at infinity fix the constant.
Step 1: Isolate $y'$ and integrate
\[y'=\frac{1+\cos2x}{x^2} \quad\Longrightarrow\quad dy=\left(\frac{1}{x^2}+\frac{\cos2x}{x^2}\right)dx.\]
The first term integrates to $-1/x$. For the second, integrate by parts with $u=\cos2x,\ dv=x^{-2}dx$ (so $du=-2\sin2x\,dx,\ v=-1/x$):
\[\int\frac{\cos2x}{x^2}\,dx=-\frac{\cos2x}{x}-2\int\frac{\sin2x}{x}\,dx=-\frac{\cos2x}{x}-2\operatorname{Si}(2x)+C_1,\]
using $\int\frac{\sin2x}{x}\,dx=\operatorname{Si}(2x)+C$.
Step 2: General solution
Combining,
\[\boxed{\,y=-\frac{1+\cos2x}{x}-2\operatorname{Si}(2x)+C.}\]
Step 3: Apply the condition at infinity
As $x\to\infty$, $\dfrac{1+\cos2x}{x}\to0$ (bounded numerator over a growing denominator) and $\operatorname{Si}(2x)\to\dfrac{\pi}{2}$, so
\[\lim_{x\to\infty}y=0-2\cdot\frac{\pi}{2}+C=C-\pi.\]
Setting this equal to $\dfrac{9\pi}{4}$ gives $C=\dfrac{9\pi}{4}+\pi=\dfrac{13\pi}{4}$.
Result
Final answer:
\[\boxed{\,y=\frac{13\pi}{4}-2\operatorname{Si}(2x)-\frac{1+\cos2x}{x},\qquad x>0.}\]