Show that y=x is a solution of the differential equation y'' - (1/x)y' + (1/x²)y = 0 and find its general solution.

Given $y = x$.
The first derivative is $y' = \frac{d}{dx}(x) = 1$.
The second derivative is $y'' = \frac{d}{dx}(1) = 0$.

Substitute $y=x$, $y'=1$, and $y''=0$ into $y'' - \frac{1}{x}y' + \frac{1}{x^2}y = 0$:

Since the equation holds true, $y=x$ is indeed a solution to the differential equation.

First, find $\int P(x) dx$:

Then, calculate $e^{-\int P(x) dx}$:

Assuming $x > 0$, we have $e^{-\int P(x) dx} = x$.

The formula for the second solution $y_2(x)$ is:

Substitute $y_1(x) = x$ and $e^{-\int P(x) dx} = x$:

Again, assuming $x > 0$, $y_2(x) = x \ln x$.

The general solution is a linear combination of the two linearly independent solutions $y_1(x)$ and $y_2(x)$:

The final answer is $\boxed{y(x) = c_1 x + c_2 x \ln x}$.