Step-by-Step Solution

1. The given differential equation is: $$\frac{dy}{dx} + \sec^{2}x \cdot y = \tan x \cdot \sec^{2}x$$

2. This is a linear differential equation of the form $$\frac{dy}{dx} + P(x)y = Q(x)$$, where $$P(x) = \sec^{2}x$$ and $$Q(x) = \tan x \cdot \sec^{2}x$$

3. The integrating factor (IF) is given by: $$IF = e^{\int P(x) dx} = e^{\int \sec^{2}x dx} = e^{\tan x}$$

4. The general solution is given by: $$y \cdot IF = \int (Q(x) \cdot IF) dx + C$$

   $$y \cdot e^{\tan x} = \int (\tan x \cdot \sec^{2}x \cdot e^{\tan x}) dx + C$$

5. Let $$u = \tan x$$, then $$\frac{du}{dx} = \sec^{2}x$$, so $$du = \sec^{2}x dx$$

   The integral becomes: $$\int u \cdot e^{u} du$$

6. Using integration by parts: $$\int u \cdot e^{u} du = u \cdot e^{u} - \int e^{u} du = u \cdot e^{u} - e^{u} + C_{1} = e^{u}(u - 1) + C_{1}$$

   Substituting back $$u = \tan x$$: $$\int \tan x \cdot \sec^{2}x \cdot e^{\tan x} dx = e^{\tan x}(\tan x - 1) + C_{1}$$

7. Therefore, the general solution is: $$y \cdot e^{\tan x} = e^{\tan x}(\tan x - 1) + C$$

   $$y = \tan x - 1 + Ce^{-\tan x}$$

8. Given the condition $$y(0) = 0$$: $$0 = \tan(0) - 1 + Ce^{-\tan(0)}$$

   $$0 = 0 - 1 + Ce^{0}$$

   $$0 = -1 + C$$

   $$C = 1$$

9. The particular solution is: $$y = \tan x - 1 + e^{-\tan x}$$