Answer
$$y(x) = e^{2-\frac{x^2}{2}}+1$$
Work Step by Step
The given integral equation is $y(x) = 2 + \int_{2}^{x} (t - ty(t)) dt$. By the fundamental theorem of calculus, we have $\frac{dy}{dx} = x - xy$. This is a separable differential equation, which can be written as $\frac{dy}{dx} = \frac{x}{1/(1-y)}$. Separating the variables and integrating both sides, we get:
$$\int \frac{1}{1-y} dy = \int x dx$$
$$-\ln(y-1) = \frac{x^2}{2} + C$$
where $C$ is a constant of integration. Solving for $y$, we get:
$$1/(y-1) = e^{\frac{x^2}{2} + C}$$
$$y = e^{C - \frac{x^2}{2}}+1$$
We can use the initial condition $y(2) = 2$ to find the value of the constant $C$. Substituting $x = 2$ and $y(2) = 2$ into the above equation, we get:
$$y(2) = e^{C - 2}+1 = 2$$
Solving for $C$, we get:
$$C = 2$$
Therefore, the solution to the given integral equation is:
$$y(x) = e^{2-\frac{x^2}{2}}+1$$
