Solve: `cos^(2)x(dy)/(dx)+y=tanx`

Answer

To solve the differential equation \( \cos^2 x \frac{dy}{dx} + y = \tan x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^2 x \frac{dy}{dx} + y = \tan x \]
We can rearrange it to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\tan x - y}{\cos^2 x} \]

### Step 2: Divide the entire equation by \( \cos^2 x \)
To simplify, we divide the whole equation by \( \cos^2 x \):
\[ \frac{dy}{dx} + y \sec^2 x = \tan x \sec^2 x \]
This is now in the standard form of a linear first-order differential equation:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \( P(x) = \sec^2 x \) and \( Q(x) = \tan x \sec^2 x \).

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \sec^2 x \, dx} \]
The integral of \( \sec^2 x \) is \( \tan x \), so:
\[ \mu(x) = e^{\tan x} \]

### Step 4: Multiply the equation by the integrating factor
Now, we multiply the entire differential equation by the integrating factor:
\[ e^{\tan x} \frac{dy}{dx} + e^{\tan x} y \sec^2 x = e^{\tan x} \tan x \sec^2 x \]

### Step 5: Recognize the left-hand side as a derivative
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(y e^{\tan x}) = e^{\tan x} \tan x \sec^2 x \]

### Step 6: Integrate both sides
Integrating both sides with respect to \( x \):
\[ \int \frac{d}{dx}(y e^{\tan x}) \, dx = \int e^{\tan x} \tan x \sec^2 x \, dx \]
The left side simplifies to:
\[ y e^{\tan x} = \int e^{\tan x} \tan x \sec^2 x \, dx + C \]

### Step 7: Substitute back and solve for \( y \)
To solve the integral on the right side, we can use substitution. Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \):
\[ \int e^{t} t \, dt \]
Using integration by parts, we find:
\[ \int t e^{t} \, dt = t e^{t} - e^{t} + C \]
Substituting back:
\[ y e^{\tan x} = \tan x e^{\tan x} - e^{\tan x} + C \]
Dividing through by \( e^{\tan x} \):
\[ y = \tan x - 1 + C e^{-\tan x} \]

### Final Answer
Thus, the solution to the differential equation is:
\[ y = \tan x - 1 + C e^{-\tan x} \]