Find the general solution of (1-x^2) y'' - x y' - 2 = 0

Find the general solution of $(1-x^2)\,y'' - x\,y' - 2 = 0$.

Step 1: Substitution

Let $p = y'$, so $p' = y''$. The equation becomes: \[(1-x^2)\,p' - x\,p = 2. \tag{1}\]

This is a first-order linear ODE in $p(x)$.

Step 2: Standard form

Divide through by $(1-x^2)$: \[p' - \frac{x}{1-x^2}\,p = \frac{2}{1-x^2}. \tag{2}\]

Step 3: Integrating factor

The integrating factor is: $\mu = e^{\displaystyle\int -\frac{x}{1-x^2}\,dx}.$

Compute the integral using $u = 1-x^2$, $du = -2x\,dx$: \[\int -\frac{x}{1-x^2}\,dx = \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|1-x^2| = \ln\sqrt{1-x^2}.\]

Thus: $\mu = \sqrt{1-x^2}.$

Step 4: Multiply by the integrating factor

Multiply (2) by $\sqrt{1-x^2}$: $\frac{d}{dx}\left[\sqrt{1-x^2}\,p\right] = \frac{2}{\sqrt{1-x^2}}.$

Step 5: Integrate both sides

\[ \sqrt{1-x^2}\,p = \int \frac{2}{\sqrt{1-x^2}}\,dx = 2\arcsin x + C_1. \]

Thus: \[p = y' = \frac{2\arcsin x + C_1}{\sqrt{1-x^2}}. \tag{3}\]

Step 6: Second integration

Integrate (3). Split into two integrals: $y = \int \frac{2\arcsin x}{\sqrt{1-x^2}}\,dx + C_1\int \frac{dx}{\sqrt{1-x^2}} + C_2.$

For the first integral, use $u = \arcsin x$, $du = \dfrac{dx}{\sqrt{1-x^2}}$: \[\int \frac{2\arcsin x}{\sqrt{1-x^2}}\,dx = 2\int u\,du = u^2 = (\arcsin x)^2.\]

For the second integral: $C_1\int \frac{dx}{\sqrt{1-x^2}} = C_1 \arcsin x.$

Step 7: General solution

The solution is recorded in the result below.

Step 8: Verification

Differentiate: \[y' = \frac{2\arcsin x}{\sqrt{1-x^2}} + \frac{C_1}{\sqrt{1-x^2}} = \frac{2\arcsin x + C_1}{\sqrt{1-x^2}},\]

\[ y'' = \frac{\dfrac{2}{\sqrt{1-x^2}}\cdot\sqrt{1-x^2} - (2\arcsin x + C_1)\cdot\dfrac{-x}{\sqrt{1-x^2}}}{1-x^2} = \frac{2 + \dfrac{x(2\arcsin x+C_1)}{\sqrt{1-x^2}}}{1-x^2}. \]

Check $(1-x^2)y'' - xy' - 2$: \[(1-x^2)\cdot\frac{2 + \dfrac{x(2\arcsin x+C_1)}{\sqrt{1-x^2}}}{1-x^2} - x\cdot\frac{2\arcsin x+C_1}{\sqrt{1-x^2}} - 2\]

\[ = 2 + \frac{x(2\arcsin x+C_1)}{\sqrt{1-x^2}} - \frac{x(2\arcsin x+C_1)}{\sqrt{1-x^2}} - 2 = 0.\ \checkmark \]