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 $\sec x$.
Answer & explanation
Correct answer: option 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,
$y(0)=1$
$⇒\sin 0+c(\cos 0)=1$
$⇒c=1$
$∴y=\sin x+\cos x$
$=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)=\sqrt{2}×\frac{1}{\sqrt{2}}\sin x+\sqrt{2}×\frac{1}{\sqrt{2}}\cos x$