Key Concepts and Formulas

• Solving Differential Equations: The general solution to a differential equation of the form $\frac{dy}{dx} = f(x)$ is found by integrating both sides with respect to $x$: $y(x) = \int f(x) \, dx + C$, where $C$ is the constant of integration.
• Trigonometric Identities: We will use the identity $\sin 2x = \frac{2\tan x}{1 + \tan^2 x}$ and the fact that $1 + \tan^2 x = \sec^2 x$. Also, we need $\tan(\frac{\pi}{4}) = 1$.
• Partial Fractions: We will need to decompose a rational function into partial fractions to evaluate an integral.
• Finding Intersections of Curves: To find the points of intersection of two curves, we set their equations equal to each other and solve for $x$.

Step-by-Step Solution

Step 1: Solve the Differential Equation

We are given the differential equation $\frac{dy}{dx} = \frac{1}{1 + \sin 2x}$. To solve for $y(x)$, we integrate both sides with respect to $x$: $y(x) = \int \frac{1}{1 + \sin 2x} \, dx + C$ To evaluate the integral, we use the substitution $t = \tan x$. Then, $\sin 2x = \frac{2t}{1 + t^2}$ and $dx = \frac{dt}{1 + t^2}$. Substituting these into the integral, we get: $y(x) = \int \frac{1}{1 + \frac{2t}{1 + t^2}} \cdot \frac{1}{1 + t^2} \, dt + C = \int \frac{1}{1 + t^2 + 2t} \, dt + C = \int \frac{1}{(1 + t)^2} \, dt + C$ Now, we can easily integrate: $y(x) = -\frac{1}{1 + t} + C = -\frac{1}{1 + \tan x} + C$

Step 2: Apply the Initial Condition

We are given the initial condition $y\left(\frac{\pi}{4}\right) = \frac{1}{2}$. Substituting $x = \frac{\pi}{4}$ into the equation for $y(x)$, we get: $\frac{1}{2} = -\frac{1}{1 + \tan\left(\frac{\pi}{4}\right)} + C = -\frac{1}{1 + 1} + C = -\frac{1}{2} + C$ Solving for $C$, we find $C = 1$. Therefore, the particular solution to the differential equation is: $y(x) = -\frac{1}{1 + \tan x} + 1 = \frac{1 + \tan x - 1}{1 + \tan x} = \frac{\tan x}{1 + \tan x}$

Step 3: Find the Intersection Points

We want to find the points of intersection between the curve $y = y(x) = \frac{\tan x}{1 + \tan x}$ and the curve $y = \sqrt{2} \sin x$. Setting the two equations equal to each other, we get: $\frac{\tan x}{1 + \tan x} = \sqrt{2} \sin x$ $\frac{\sin x}{\cos x (1 + \frac{\sin x}{\cos x})} = \sqrt{2} \sin x$ $\frac{\sin x}{\cos x + \sin x} = \sqrt{2} \sin x$ If $\sin x = 0$, then $x = 0, \pi, 2\pi$. However, $x \in S = (0, 2\pi) - \left\{ \frac{\pi}{2}, \frac{3\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4} \right\}$. Thus, $x = \pi$ is a possible solution.

If $\sin x \neq 0$, we can divide both sides by $\sin x$: $\frac{1}{\cos x + \sin x} = \sqrt{2}$ $\cos x + \sin x = \frac{1}{\sqrt{2}}$ Multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$, we get $\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{1}{2}$ $\cos\left(\frac{\pi}{4}\right) \cos x + \sin\left(\frac{\pi}{4}\right) \sin x = \frac{1}{2}$ $\cos\left(x - \frac{\pi}{4}\right) = \frac{1}{2}$ Therefore, $x - \frac{\pi}{4} = \frac{\pi}{3}$ or $x - \frac{\pi}{4} = -\frac{\pi}{3}$. This gives $x = \frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12}$ and $x = \frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$. Since $x \in (0, 2\pi)$, $x = 2\pi - \frac{\pi}{12} = \frac{23\pi}{12}$. Therefore, the solutions are $x = \pi, \frac{7\pi}{12}, \frac{23\pi}{12}$.

Step 4: Sum the Abscissas

The sum of the abscissas of the points of intersection is: $\pi + \frac{7\pi}{12} + \frac{23\pi}{12} = \pi + \frac{30\pi}{12} = \pi + \frac{5\pi}{2} = \frac{7\pi}{2} = \frac{42\pi}{12}$ We are given that this sum is equal to $\frac{k\pi}{12}$. Therefore, $\frac{k\pi}{12} = \frac{42\pi}{12}$, which implies $k = 42$.

Common Mistakes & Tips

• Domain Restrictions: Always check that your solutions fall within the specified domain. In this case, $x \in S$.
• Trigonometric Identities: Be comfortable with manipulating trigonometric identities to simplify expressions.
• Checking for extraneous solutions: When dividing by a trigonometric function like $\sin x$, remember to consider the case where $\sin x = 0$ separately.

Summary

We solved the given differential equation using trigonometric substitution and partial fractions. We applied the initial condition to find the particular solution. We then found the intersection points of the solution curve with $y = \sqrt{2} \sin x$ by setting the equations equal to each other and solving for $x$. Finally, we summed the abscissas of the intersection points and equated the result to $\frac{k\pi}{12}$ to find the value of $k$.

Final Answer

The final answer is \boxed{42}.