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 ______

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

Since $$1 + \sin 2x = 1 + 2\sin x\cos x = \sin^2 x + \cos^2 x + 2\sin x\cos x = (\sin x + \cos x)^2$$, it follows that $$\frac{dy}{dx} = \frac{1}{(\sin x + \cos x)^2}$$.

Writing $$\sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$$ gives $$\frac{1}{(\sin x + \cos x)^2} = \frac{1}{2\sin^2\left(x + \frac{\pi}{4}\right)} = \frac{1}{2}\csc^2\left(x + \frac{\pi}{4}\right)$$, so that $$y = \int \frac{1}{2}\csc^2\left(x + \frac{\pi}{4}\right)\,dx = -\frac{1}{2}\cot\left(x + \frac{\pi}{4}\right) + C$$.

Applying the initial condition yields $$y\left(\frac{\pi}{4}\right) = -\frac{1}{2}\cot\left(\frac{\pi}{2}\right) + C = 0 + C = C$$, and hence $$C = \frac{1}{2}$$. Therefore $$y = \frac{1}{2} - \frac{1}{2}\cot\left(x + \frac{\pi}{4}\right) = \frac{1}{2}\left(1 - \cot\left(x + \frac{\pi}{4}\right)\right)$$.

Now rewriting this in terms of $$\tan$$ gives $$y = \frac{1}{2}\left(1 - \frac{\cos\left(x + \frac{\pi}{4}\right)}{\sin\left(x + \frac{\pi}{4}\right)}\right) = \frac{1}{2}\cdot\frac{\sin\left(x + \frac{\pi}{4}\right) - \cos\left(x + \frac{\pi}{4}\right)}{\sin\left(x + \frac{\pi}{4}\right)}$$. Using the sum formulas leads to $$\sin\left(x + \frac{\pi}{4}\right) - \cos\left(x + \frac{\pi}{4}\right) = \frac{\sin x + \cos x}{\sqrt{2}} - \frac{\cos x - \sin x}{\sqrt{2}} = \frac{2\sin x}{\sqrt{2}} = \sqrt{2}\sin x$$, and hence $$y = \frac{\sqrt{2}\sin x}{2\sin\left(x + \frac{\pi}{4}\right)} = \frac{\sqrt{2}\sin x}{2 \cdot \frac{1}{\sqrt{2}}(\sin x + \cos x)} = \frac{2\sin x}{2(\sin x + \cos x)} = \frac{\sin x}{\sin x + \cos x}$$.

Next, to find the intersection with $$y = \sqrt{2}\sin x$$ we solve $$\frac{\sin x}{\sin x + \cos x} = \sqrt{2}\sin x$$. If $$\sin x = 0$$ then $$x = \pi$$ in $$(0,2\pi)$$. Otherwise dividing by $$\sin x$$ gives $$\frac{1}{\sin x + \cos x} = \sqrt{2}$$ so that $$\sin x + \cos x = \frac{1}{\sqrt{2}}$$. Since $$\sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$$ this implies $$\sin\left(x + \frac{\pi}{4}\right) = \frac{1}{2}$$ and hence $$x + \frac{\pi}{4} = \frac{\pi}{6}, \frac{5\pi}{6}, \text{ or } \frac{\pi}{6} + 2\pi$$, giving $$x = -\frac{\pi}{12}, \frac{7\pi}{12}, \frac{23\pi}{12}$$. In $$(0,2\pi)$$ the valid solutions are $$x = \frac{7\pi}{12}$$ and $$x = \frac{23\pi}{12}$$ since these are not in $$\{\frac{\pi}{2},\frac{3\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}\}$$. Moreover, checking $$x = \pi$$ yields $$y(\pi) = \frac{0}{-1} = 0$$ and $$\sqrt{2}\sin\pi = 0$$, so that $$x = \pi$$ also works.

Finally, summing the abscissas gives $$\frac{7\pi}{12} + \pi + \frac{23\pi}{12} = \frac{7\pi + 12\pi + 23\pi}{12} = \frac{42\pi}{12} = \frac{k\pi}{12}$$, so that $$k = 42$$. The answer is $$\boxed{42}$$.