Prove that y = (arcsin x)² satisfies (1-x²)y''-xy' = 2
Prove that $y = \bigl(\arcsin x\bigr)^{2}$ satisfies $\bigl(1-x^{2}\bigr)y''-xy' = 2$.
Explanation of Method
Clearing the radical before the second differentiation.
Step 1: First derivative
\[ y' = 2\arcsin x\cdot\frac{1}{\sqrt{1-x^{2}}} = \frac{2\arcsin x}{\sqrt{1-x^{2}}}. \]
Step 2: Clear the radical first
Differentiating this quotient directly would be unpleasant. Instead, move the root to the left: \[y'\sqrt{1-x^{2}} = 2\arcsin x. \tag{1}\]
Now both sides are easy to differentiate, and the arcsine will disappear on the right.
Step 3: Differentiate (1)
Product rule on the left, with $\bigl(\sqrt{1-x^{2}}\bigr)' = \dfrac{-x}{\sqrt{1-x^{2}}}$: $y''\sqrt{1-x^{2}}-\frac{xy'}{\sqrt{1-x^{2}}} = \frac{2}{\sqrt{1-x^{2}}}$.
Step 4: Multiply through by $\sqrt{1-x^{2}}$
\[ \bigl(1-x^{2}\bigr)y''-xy' = 2, \]
which is the required identity.
Step 5: Verification
Differentiating $y'$ explicitly by the quotient rule gives
\[y'' = \frac{\dfrac{2}{\sqrt{1-x^{2}}}\sqrt{1-x^{2}}-2\arcsin x\cdot\dfrac{-x}{\sqrt{1-x^{2}}}}{1-x^{2}} = \frac{2}{1-x^{2}}+\frac{2x\arcsin x}{\bigl(1-x^{2}\bigr)^{3/2}}\]
\[\bigl(1-x^{2}\bigr)y''-xy' = 2+\frac{2x\arcsin x}{\sqrt{1-x^{2}}}-\frac{2x\arcsin x}{\sqrt{1-x^{2}}} = 2 \;\;\checkmark\]
\[x = 0:\ y(0) = 0,\ y'(0) = 0, \text{ and the equation gives } y''(0) = 2 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\bigl(1-x^{2}\bigr)y''-xy' = 2 \quad\text{holds on } |x|\lt 1\,}\]
Remark
Remark. The equation is the reason $(\arcsin x)^{2}$ has such a clean Maclaurin series. Differentiating $\bigl(1-x^{2}\bigr)y''-xy' = 2$ repeatedly and setting $x = 0$ yields the recurrence $y^{(n+2)}(0) = n^{2}y^{(n)}(0)$, from which $y^{(2k)}(0) = 2\bigl[2\cdot4\cdots(2k-2)\bigr]^{2}$ and all odd derivatives vanish — giving $\bigl(\arcsin x\bigr)^{2} = x^{2}+\tfrac13 x^{4}+\tfrac{8}{45}x^{6}+\cdots$.