We are given the following differential equation to solve:

$y' \cdot \sin x - (2y + 1) \cos x = 0$ with the initial condition: $y\left(\frac{\pi}{3}\right) = 1$

We need to find the general solution of the equation and choose the correct solution from the given options.

Step 1: Rearrange the differential equation

First, we can rewrite the given equation as follows:

$y' \cdot \sin x = (2y + 1) \cos x$

Next, divide both sides by $\sin x$ (assuming $\sin x \neq 0$):

$y' = (2y + 1) \frac{\cos x}{\sin x} = (2y + 1) \cot x$

This is a first-order linear differential equation.

Step 2: Solve the differential equation using the integrating factor

The equation is of the form:

$y' - (2 \cot x)y = \cot x$

This is a linear first-order differential equation. To solve it, we will first find an integrating factor. The integrating factor $\mu(x)$ is given by:

$\mu(x) = e^{\int -2 \cot x \, dx} = e^{-2 \ln |\sin x|} = \frac{1}{\sin^2 x}$

Step 3: Multiply the equation by the integrating factor

Multiplying the entire differential equation by $\frac{1}{\sin^2 x}$:

$\frac{1}{\sin^2 x} y' - \frac{2}{\sin^2 x} \cot x \, y = \frac{1}{\sin^2 x} \cot x$

The left-hand side simplifies to:

$\frac{d}{dx} \left( \frac{y}{\sin^2 x} \right) = \frac{\cot x}{\sin^2 x}$

Step 4: Integrate both sides

Now, integrate both sides with respect to $x$:

$\frac{y}{\sin^2 x} = \int \frac{\cot x}{\sin^2 x} \, dx$

We can express $\frac{\cot x}{\sin^2 x}$ as $\frac{\cos x}{\sin^3 x}$. This is a standard integral, and its result is:

$\int \frac{\cos x}{\sin^3 x} \, dx = -\frac{1}{2 \sin^2 x} + C$

Thus, we have:

$\frac{y}{\sin^2 x} = -\frac{1}{2 \sin^2 x} + C$

Step 5: Solve for $y$

Multiply both sides by $\sin^2 x$:

$y = -\frac{1}{2} + C \sin^2 x$

Step 6: Apply the initial condition

We are given that $y\left(\frac{\pi}{3}\right) = 1$. Substituting $x = \frac{\pi}{3}$ into the equation:

$1 = -\frac{1}{2} + C \sin^2\left(\frac{\pi}{3}\right)$

Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$, we get:

$1 = -\frac{1}{2} + C \left(\frac{\sqrt{3}}{2}\right)^2 = -\frac{1}{2} + C \cdot \frac{3}{4}$

Solving for $C$:

$1 + \frac{1}{2} = C \cdot \frac{3}{4}$ $\frac{3}{2} = C \cdot \frac{3}{4}$ $C = 2$

Final Solution:

Substitute $C = 2$ into the general solution:

$y = -\frac{1}{2} + 2 \sin^2 x$ Simplify the expression:

$y = 2 \sin^2 x - \frac{1}{2}$

Multiply everything by 2 to make the equation easier to compare with the options:

$2y + 1 = 4 \sin^2 x$

Taking the square root of both sides:

$\sqrt{2y + 1} = 2 \sin x$

Thus, the correct answer is:

$\boxed{\sqrt{y + 1} = 2 \sin x}$

Would you like any further details on this solution?

Related Questions:

1. What are some common techniques for solving first-order linear differential equations?
2. How do you determine the integrating factor for a linear differential equation?
3. What is the method of separation of variables in solving differential equations?
4. How can initial conditions help determine the constant of integration?
5. What is the significance of the Cauchy problem in differential equations?

Tip:

When solving differential equations, always check whether you can simplify by using known identities, such as trigonometric identities or logarithmic properties.