If $\cos x \frac{\mathrm{~d} y}{\mathrm{~d} x}-y \sin x=6 x, 0 \lt x \lt \frac{\pi}{2}$, then general…
If $\cos x \frac{\mathrm{~d} y}{\mathrm{~d} x}-y \sin x=6 x, 0 \lt x \lt \frac{\pi}{2}$, then general solution of the differential equation is
$y=\cos x+3 x^2+\mathrm{c}$, where c is a constant of integration.
$y+\cos x=3 x^2+\mathrm{c}$, where c is a constant of integration.
$y=3 x^2 \cos x+\cos x$, where c is a constant of integration.
$y \cdot \cos x=3 x^2+\mathrm{c}$, where c is a constant of integration.
Solution
$\begin{aligned}
& \cos x \frac{\mathrm{~d} y}{\mathrm{~d} x}-y \sin x=6 x \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}-(\tan x) y=6 x \sec x
\end{aligned}$
This equation is of the form $\frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{p} y=\mathrm{Q}$
$\therefore \quad \text { If }=\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{-\int \tan x d x}=\mathrm{e}^{\log x \cos x}=\cos x$
$\therefore \quad$ Solution of given equation is
$\begin{aligned}
& y \cdot \cos x=\int 6 x \times \sec x \times \cos x \mathrm{~d} x+\mathrm{c} \\
& \Rightarrow y \cos x=\int 6 x \mathrm{~d} x+\mathrm{c} \\
& \Rightarrow y \cos x=3 x^2+\mathrm{c}
\end{aligned}$