Find the general solution of the differential equation: `cos^(2)x(dy)/(dx)+y=tanx`. Find the particular solution which satisfies `y=0` at `x=0`.

Answer

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

### Step 1: Rearranging the Equation
We start by rewriting the equation in a standard form:
\[ \frac{dy}{dx} + \frac{y}{\cos^2 x} = \frac{\tan x}{\cos^2 x} \]
This simplifies to:
\[ \frac{dy}{dx} + y \sec^2 x = \tan x \sec^2 x \]

### Step 2: Identifying \( p(x) \) and \( q(x) \)
Here, we identify:
- \( p(x) = \sec^2 x \)
- \( q(x) = \tan x \sec^2 x \)

### Step 3: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \sec^2 x \, dx} = e^{\tan x} \]

### Step 4: Multiplying the Equation by the Integrating Factor
We multiply the entire differential equation by the integrating factor:
\[ e^{\tan x} \frac{dy}{dx} + e^{\tan x} y \sec^2 x = e^{\tan x} \tan x \sec^2 x \]
This simplifies to:
\[ \frac{d}{dx}(e^{\tan x} y) = e^{\tan x} \tan x \sec^2 x \]

### Step 5: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{d}{dx}(e^{\tan x} y) \, dx = \int e^{\tan x} \tan x \sec^2 x \, dx \]
The left side simplifies to:
\[ e^{\tan x} y = \int e^{\tan x} \tan x \sec^2 x \, dx \]

### Step 6: Solving the Right Side Integral
To solve the right side, we can use integration by parts. Let:
- \( u = \tan x \)
- \( dv = e^{\tan x} \sec^2 x \, dx \)

Then, we have:
- \( du = \sec^2 x \, dx \)
- \( v = e^{\tan x} \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives:
\[ \int \tan x e^{\tan x} \sec^2 x \, dx = \tan x e^{\tan x} - \int e^{\tan x} \sec^2 x \, dx \]

### Step 7: Final Form of the General Solution
After solving the integral, we find:
\[ e^{\tan x} y = e^{\tan x} \tan x - e^{\tan x} + C \]
Dividing by \( e^{\tan x} \):
\[ y = \tan x - 1 + Ce^{-\tan x} \]

### Step 8: Finding the Particular Solution
To find the particular solution that satisfies \( y(0) = 0 \):
Substituting \( x = 0 \):
\[ 0 = \tan(0) - 1 + Ce^{-\tan(0)} \]
This simplifies to:
\[ 0 = 0 - 1 + C \cdot 1 \implies C = 1 \]

### Final Particular Solution
Substituting \( C = 1 \) back into the general solution:
\[ y = \tan x - 1 + e^{-\tan x} \]

### Summary
The general solution is:
\[ y = \tan x - 1 + Ce^{-\tan x} \]
The particular solution is:
\[ y = \tan x - 1 + e^{-\tan x} \]