If y = (sin^(-1)(x))^2, prove that (1-x^2)y'' - xy' - 2 = 0.

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Step-by-Step Solution

Step 1: Differentiate with respect to x

We begin by differentiating the given equation y = (\sin^{-1}(x))^2 with respect to x$. We use the chain rule, where the derivative of u^n is nu^{n-1} \frac{du}{dx} and the derivative of \sin^{-1}(x) is \frac{1}{\sqrt{1-x^2}}$.

Step 2: Rearrange and square both sides

To simplify the expression and prepare for the second differentiation, we multiply both sides by $\sqrt{1-x^2}$. Then, we square both sides to eliminate the square root. This step also allows us to substitute y back into the equation.

Step 3: Substitute y and differentiate again

Now, we substitute y = (\sin^{-1}(x))^2 back into the equation. Then, we differentiate both sides with respect to x again. We use the product rule for the left side and the chain rule for the right side.

Step 4: Simplify the equation

We simplify the equation by dividing all terms by 2y'$. This is valid as long as y' eq 0$. Since y = (\sin^{-1}(x))^2$, y' is zero only at x=0$, which is a specific point, not a general case. Rearranging the terms gives us the desired result.