The differential equation $\mathrm{dy}=(\mathrm{y} \tan \mathrm{x}+\sec \mathrm{x}) \mathrm{d} \mathrm{x}$ can be written as
$\frac{d y}{d x}=y \tan x+\sec x$
or, $\quad \frac{d y}{d x}-y \tan x=\sec x$
which is of the form $\frac{d y}{d x}+P(x) \cdot y=Q(x)$ Here $\mathrm{P}(\mathrm{x})=-\tan \mathrm{x}$ and $\mathrm{Q}(\mathrm{x})=\sec \mathrm{x}$
Integrating factor $\mathrm{IF}=\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}$
$I F=e^{\int-\tan x d x}=e^{\int-\frac{\sin x}{\cos x} d x}$
Putting $\cos \mathrm{x}=\mathrm{t}$
$-\sin x d x=d t$
$I f=\mathrm{e}^{\int \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{\log _{\mathrm{e}} \mathrm{t}}=\mathrm{t}=\cos \mathrm{x}$
The solution is
$\mathrm{y} \cdot \mathrm{Q}(\mathrm{x})=\int \mathrm{I} \mathrm{F} \cdot \mathrm{Q}(\mathrm{x}) \mathrm{d} \mathrm{x}+\mathrm{c}$
or, $\mathrm{y} \cdot \sec \mathrm{x}=\int \cos \mathrm{x} \cdot \sec \mathrm{x} \mathrm{d} \mathrm{x}+\mathrm{c}$
or, $\mathrm{y} \cdot \sec \mathrm{x}=\int \mathrm{d} \mathrm{x}+\mathrm{c}$
or, $y \cdot \sec x=x+c$
Since the curve passes through the origin. $0=0+\mathrm{c} \Rightarrow \mathrm{c}=0$
and $y \sec x=x$
or, $\mathrm{y}=\mathrm{x} \cos \mathrm{x}$