Let a solution $y=y(x)$ of the differential equation $\frac{d y}{d x} \cos x+y \sin x=1$ satisfy $y(0)=1$.

Statement-1: $y(x)=\sqrt{2}\sin \left(\frac{\pi}{4}+x\right)$

Statement-2: The integrating factor of the given differential equation is $\sec x$.

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

We have,

$\frac{d y}{d x} \cos x+y \sin x=1$

$\Rightarrow \frac{d y}{d x}+y \tan x=\sec x$

$I.F.=e^{\int\tan xdx}=e^{\log|\sec x|}$

$=\sec x$

$∴y\sec x=\int\sec^2xdx$

$⇒y\sec x=\tan x+c$

$⇒y=\frac{\tan x+c}{\sec x}$

$⇒y=\sin x+c\cos x$

and,

$y(0)=1$

$⇒\sin 0+c(\cos 0)=1$

$⇒c=1$

$∴y=\sin x+\cos x$

$=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)=\sqrt{2}×\frac{1}{\sqrt{2}}\sin x+\sqrt{2}×\frac{1}{\sqrt{2}}\cos x$