Solve the differential equation x² y'' - xy' + y = 0 if y = 1/x is a solution

The given differential equation is $x^2 y'' - 2y = 0$.
To use the reduction of order formula, we need the equation in the form $y'' + P(x)y' + Q(x)y = 0$.
Divide the entire equation by $x^2$:

$$y'' - \frac{2}{x^2}y = 0$$

From this, we can identify $P(x) = 0$ and $Q(x) = -\frac{2}{x^2}$.
The given solution is $y_1(x) = \frac{1}{x}$.

Since $P(x) = 0$:

$$\int P(x) dx = \int 0 \, dx = 0$$

Therefore,

$$e^{-\int P(x) dx} = e^{-0} = 1$$

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

$$y_2(x) = y_1(x) \int \frac{e^{-\int P(x) dx}}{(y_1(x))^2} dx$$

Substitute $y_1(x) = \frac{1}{x}$ and $e^{-\int P(x) dx} = 1$:

$$y_2(x) = \frac{1}{x} \int \frac{1}{\left(\frac{1}{x}\right)^2} dx$$ $$y_2(x) = \frac{1}{x} \int \frac{1}{\frac{1}{x^2}} dx$$ $$y_2(x) = \frac{1}{x} \int x^2 dx$$

$$\int x^2 dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$$

Now substitute this back into the expression for $y_2(x)$:

$$y_2(x) = \frac{1}{x} \left(\frac{x^3}{3}\right)$$ $$y_2(x) = \frac{x^2}{3}$$

We can absorb the constant $\frac{1}{3}$ into the arbitrary constant $c_2$ in the general solution, so we can use $y_2(x) = x^2$.

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

$$y(x) = c_1 y_1(x) + c_2 y_2(x)$$ $$y(x) = c_1 \left(\frac{1}{x}\right) + c_2 (x^2)$$ $$y(x) = \frac{c_1}{x} + c_2 x^2$$