We are given the differential equation:

$\cos(x) \frac{dy}{dx} + \sin(x) y = 1$

To solve this first-order linear differential equation, we'll use an integrating factor approach.

Step 1: Rewrite the equation in standard form

We start by getting the equation into the form:

$\frac{dy}{dx} + P(x) y = Q(x)$

Dividing through by $\cos(x)$, we get:

$\frac{dy}{dx} + \frac{\sin(x)}{\cos(x)} y = \frac{1}{\cos(x)}$

This simplifies to:

$\frac{dy}{dx} + \tan(x) y = \sec(x)$

Now, the equation is in the standard form with:

• $P(x) = \tan(x)$
• $Q(x) = \sec(x)$

Step 2: Find the integrating factor

The integrating factor is given by:

$\mu(x) = e^{\int P(x) dx} = e^{\int \tan(x) dx}$

The integral of $\tan(x)$ is $\ln|\sec(x)|$, so the integrating factor becomes:

$\mu(x) = e^{\ln|\sec(x)|} = \sec(x)$

Step 3: Multiply through by the integrating factor

Now multiply the entire differential equation by $\sec(x)$:

$\sec(x) \frac{dy}{dx} + \sec(x) \tan(x) y = \sec^2(x)$

This simplifies to:

$\frac{d}{dx} [y \sec(x)] = \sec^2(x)$

Step 4: Integrate both sides

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

$y \sec(x) = \int \sec^2(x) dx$

The integral of $\sec^2(x)$ is $\tan(x)$, so:

$y \sec(x) = \tan(x) + C$

where $C$ is the constant of integration.

Step 5: Solve for $y$

Finally, solve for $y$:

$y = \cos(x) (\tan(x) + C)$

Simplifying:

$y = \sin(x) + C \cos(x)$

Final Solution:

The solution to the differential equation is:

$y(x) = \sin(x) + C \cos(x)$

Would you like further details or have any questions?

Here are five related questions that expand upon this concept:

1. How do you solve a second-order linear differential equation?
2. What are some real-life applications of differential equations?
3. Can we use a different method to solve this equation, such as variation of parameters?
4. How does the integrating factor method differ from separation of variables?
5. Can you solve a non-homogeneous differential equation using the same approach?

Tip: Always verify that the solution satisfies the original differential equation by substituting it back in!