Question

Let $S =(0,2 \pi)-\left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}$. Let $y =$ $y(x), x \in S$, be the solution curve of the differential equation $\frac{d y}{d x}=\frac{1}{1+\sin 2 x}, y\left(\frac{\pi}{4}\right)=\frac{1}{2}$. if the sum of abscissas of all the points of intersection of the curve $y = y ( x )$ with the curve $y =\sqrt{2} \sin x$ is $\frac{ k \pi}{12}$, then $k$ is equal to________.

Answer 1

$\frac{d y}{d x}=\frac{1}{1+\sin 2 x}$
$\int d y=\int \frac{d x}{(\sin x+\cos x)^{2}}$
$\int d y=\int \frac{\sec ^{2} x}{\left(1+\tan _{x}\right)^{2}}$
$y(x)=-\frac{1}{1+\tan x}+C$
$y\left(\frac{\pi}{4}\right)=\frac{1}{2}=-\frac{1}{2}+C$
$C=1$
$y(x)=\frac{-1}{1+\tan x}+1$
$y(x)=\frac{-1+1+\tan x}{1+\tan x}$
$y(x)=\frac{\tan x}{1+\tan x}$
Solving with $y =\sqrt{2} \sin x$
$\frac{\tan x}{1+\tan x}=\sqrt{2} \sin x$
$\sin x=0, \frac{1}{\sqrt{2}}=\sin x+\cos x$
$x=\pi \frac{1}{2}=\sin \left(x+\frac{\pi}{4}\right)$
$\sin \frac{\pi}{6}=\sin \left(x+\frac{\pi}{4}\right)$
$x+\frac{\pi}{4}=\pi-\frac{\pi}{6}, 2 \pi+\frac{\pi}{6}$
$x=\frac{5 \pi}{6}-\frac{\pi}{4}, x=\frac{13 \pi}{6}-\frac{\pi}{4}$
$x=\frac{7 \pi}{12}, x=\frac{23 \pi}{12}$
sum of sol.
$=\pi+\frac{7 \pi}{12}+\frac{23 \pi}{12}$
$=\frac{12 \pi+7 \pi+23}{12}=\frac{42 \pi}{12}=\frac{ k \pi}{12}$
$\Rightarrow k=42$