Solve the differential equation (1-x²)y'' - xy' - 6x = 0

The given differential equation is

$$(1-x^2)y'' - xy' = 6x$$

This is a second-order linear differential equation. We can reduce its order by letting $p = y'$. Then $p' = y''$.
Substituting these into the equation gives a first-order linear differential equation in terms of $p(x)$:

$$(1-x^2)p' - xp = 6x$$

Divide by $(1-x^2)$ to get it into the standard form $p' + Q(x)p = R(x)$:

$$p' - \frac{x}{1-x^2}p = \frac{6x}{1-x^2}$$

The integrating factor $\mu(x)$ is given by $e^{\int Q(x) dx}$. Here, $Q(x) = -\frac{x}{1-x^2}$.
First, calculate the integral of $Q(x)$:

$$\int -\frac{x}{1-x^2} dx$$

Let $u = 1-x^2$, then $du = -2x dx$, so $x dx = -\frac{1}{2} du$.

$$\int -\frac{x}{1-x^2} dx = \int \frac{1}{1-x^2} \left(\frac{1}{2} du\right) = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|1-x^2|$$

Assuming $1-x^2 > 0$ (i.e., $x \in (-1, 1)$), we have $\frac{1}{2} \ln(1-x^2) = \ln(\sqrt{1-x^2})$.
So, the integrating factor is:

$$\mu(x) = e^{\ln(\sqrt{1-x^2})} = \sqrt{1-x^2}$$

Multiply the standard form of the differential equation for $p$ by the integrating factor $\mu(x)$:

$$\sqrt{1-x^2} \left(p' - \frac{x}{1-x^2}p\right) = \sqrt{1-x^2} \left(\frac{6x}{1-x^2}\right)$$

The left side is the derivative of the product $p(x)\mu(x)$:

$$\frac{d}{dx}(p\sqrt{1-x^2}) = \frac{6x}{\sqrt{1-x^2}}$$

Now, integrate both sides with respect to $x$:

$$p\sqrt{1-x^2} = \int \frac{6x}{\sqrt{1-x^2}} dx$$

To evaluate the integral on the right side, again let $u = 1-x^2$, so $x dx = -\frac{1}{2} du$:

$$\int \frac{6x}{\sqrt{1-x^2}} dx = \int \frac{6}{\sqrt{u}} \left(-\frac{1}{2} du\right) = -3 \int u^{-1/2} du$$ $$= -3 \left(\frac{u^{1/2}}{1/2}\right) + C_1 = -6\sqrt{u} + C_1 = -6\sqrt{1-x^2} + C_1$$

So, we have:

$$p\sqrt{1-x^2} = -6\sqrt{1-x^2} + C_1$$

Divide by $\sqrt{1-x^2}$ to solve for $p$:

$$p = -6 + \frac{C_1}{\sqrt{1-x^2}}$$

Recall that $p = y'$, so we have:

$$y' = -6 + \frac{C_1}{\sqrt{1-x^2}}$$

Integrate $y'$ with respect to $x$ to find $y$:

$$y = \int \left(-6 + \frac{C_1}{\sqrt{1-x^2}}\right) dx$$ $$y = \int -6 dx + C_1 \int \frac{1}{\sqrt{1-x^2}} dx$$

The integral of $\frac{1}{\sqrt{1-x^2}}$ is $\arcsin(x)$.

$$y = -6x + C_1 \arcsin(x) + C_2$$

where $C_1$ and $C_2$ are arbitrary constants of integration. This solution is valid for $x \in (-1, 1)$.