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 $\frac{dy}{dx}=\frac{1}{1+\sin 2x},y\left(\frac{\pi }{4}\right)=\frac{1}{2}$. 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

Answer: 42

$\frac{dy}{dx}=\frac{1}{1+\sin 2x}$
$\int dy=\int \frac{dx}{(\sin x+\cos x{)}^2}$
$\int dy=\int \frac{{\sec }^{2}x}{{\left(1+{\tan }_x\right)}^2}$
$y(x)=-\frac{1}{1+\tan x}+C$
$y\left(\frac{\pi }{4}\right)=\frac{1}{2}=-\frac{1}{2}+C$
$C=1$
$y(x)=\frac{-1}{1+\tan x}+1$
$y(x)=\frac{-1+1+\tan x}{1+\tan x}$
$y(x)=\frac{\tan x}{1+\tan x}$
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.
$=\pi +\frac{7\pi }{12}+\frac{23\pi }{12}$
$=\frac{12\pi +7\pi +23}{12}=\frac{42\pi }{12}=\frac{k\pi }{12}$
$\Rightarrow k=42$