Let $$S = (0,2\pi ) - \left\{ {{\pi \over 2},{{3\pi } \over 4},{{3\pi } \over 2},{{7\pi } \over 4}} \right\}$$. Let $y = y(x)$, x $\in$ S, be the solution curve of the differential equation $${{dy} \over {dx}} = {1 \over {1 + \sin 2x}},\,y\left( {{\pi \over 4}} \right) = {1 \over 2}$$. 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 ${{k\pi } \over {12}}$, then k is equal to _____________.

<p>${{dy} \over {dx}} = {1 \over {1 + \sin 2x}}$</p> <p>$\Rightarrow dy = {{{{\sec }^2}xdx} \over {{{(1 + \tan x)}^2}}}$</p> <p>$\Rightarrow y = - {1 \over {1 + \tan x}} + c$</p> <p>When $x = {\pi \over 4}$, $y = {1 \over 2}$ gives c = 1</p> <p>So $$y = {{\tan x} \over {1 + \tan x}} \Rightarrow y = {{\sin x} \over {\sin x + \cos x}}$$</p> <p>Now, $y = \sqrt 2 \sin x \Rightarrow \sin x = 0$</p> <p>or $\sin x + \cos x = {1 \over {\sqrt 2 }}$</p> <p>$\sin x = 0$ gives $x = \pi$ only.</p> <p>and $$\sin x + \cos x = {1 \over {\sqrt 2 }} \Rightarrow \sin \left( {x + {\pi \over 4}} \right) = {1 \over 2}$$</p> <p>So $x + {\pi \over 4} = {{5\pi } \over 6}$ or ${{13\pi } \over 6} \Rightarrow x = {{7\pi } \over {12}}$ or ${{23\pi } \over {12}}$</p> <p>Sum of all solutions $= \pi + {{7\pi } \over {12}} + {{23\pi } \over {12}} = {{42\pi } \over {12}}$</p> <p>Hence, $k = 42$.</p>