The equation in the image is:

$(1 + \cos x) y' = (1 - e^{-x}) \sin x, \quad y(0) = 0$

This is a first-order linear differential equation. To solve it, we will follow these steps:

1. Rewrite the equation in standard linear form

The equation can be written as:

$y' + P(x)y = Q(x)$

In this case, divide both sides by $(1 + \cos x)$ to get:

$y' = \frac{(1 - e^{-x}) \sin x}{1 + \cos x}$

So, in standard form:

$y' + 0 \cdot y = \frac{(1 - e^{-x}) \sin x}{1 + \cos x}$

Here, $P(x) = 0$ and $Q(x) = \frac{(1 - e^{-x}) \sin x}{1 + \cos x}$.

2. Integrating Factor

Since $P(x) = 0$, the integrating factor is:

$e^{\int 0 \, dx} = 1$

3. General Solution

The general solution of a first-order linear differential equation is:

$y(x) = \frac{1}{\mu(x)} \left( \int \mu(x) Q(x) \, dx + C \right)$

Here, $\mu(x) = 1$, so:

$y(x) = \int Q(x) \, dx + C$

Thus, we need to integrate:

$y(x) = \int \frac{(1 - e^{-x}) \sin x}{1 + \cos x} \, dx + C$

4. Solving the Integral

Let's first break the integral into two parts:

$y(x) = \int \frac{\sin x}{1 + \cos x} \, dx - \int \frac{e^{-x} \sin x}{1 + \cos x} \, dx + C$

Part 1: $\int \frac{\sin x}{1 + \cos x} \, dx$

Use the substitution $u = 1 + \cos x$, then $du = -\sin x \, dx$:

$\int \frac{\sin x}{1 + \cos x} \, dx = -\int \frac{du}{u} = -\ln |u| = -\ln |1 + \cos x|$

Part 2: $\int \frac{e^{-x} \sin x}{1 + \cos x} \, dx$

This integral is more complex and might not have a simple closed-form solution, but for this equation, we only need the solution at specific points.

5. Apply Initial Condition

The initial condition is $y(0) = 0$.

Substitute $x = 0$ into the solution to find $C$:

$y(0) = -\ln |1 + \cos(0)| + \text{(value of the second integral at } x = 0\text{)} + C = 0$

Given that $\cos(0) = 1$, $\ln |1 + 1| = \ln 2$.

However, solving for $C$ precisely depends on the form of the second integral, but the structure is usually simplified for initial value problems in cases like these.

Thus, after simplifications, a potential simplified solution could involve finding specific values or utilizing an approximation, but the full solution would require either more detailed integration or numerical methods for the second term.

If you'd like the fully detailed expression or have any questions, feel free to ask!

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Questions:

1. What is the general form of a first-order linear differential equation?
2. How do you find the integrating factor for a differential equation?
3. Why do we divide the differential equation by $(1 + \cos x)$?
4. What is the purpose of the initial condition $y(0) = 0$?
5. What technique is used to integrate $\int \frac{\sin x}{1 + \cos x} \, dx$?
6. Why is the second integral in this problem more complex to solve?
7. How can you approximate the solution if an integral is difficult to solve analytically?
8. What are the steps to solving a differential equation with a specific initial condition?

Tip:

When dealing with integrals that are difficult to solve, always consider possible substitutions or numerical methods to approximate solutions if a closed form is not easily obtainable.