If y = sinh(ksin⁻¹x), show that (1-x²)y'' - xy' - k²y = 0

Given $y = \sinh(k \sin^{-1} x)$.
We use the chain rule. Let $u = k \sin^{-1} x$. Then $y = \sinh(u)$.
The derivative of $\sinh(u)$ with respect to $u$ is $\cosh(u)$.
The derivative of $k \sin^{-1} x$ with respect to $x$ is $k \cdot \frac{1}{\sqrt{1-x^2}}$.
So, $y' = \frac{dy}{dx} = \frac{d}{du}(\sinh(u)) \cdot \frac{du}{dx} = \cosh(k \sin^{-1} x) \cdot \frac{k}{\sqrt{1-x^2}}$.

$$y' = \frac{k \cosh(k \sin^{-1} x)}{\sqrt{1-x^2}}$$

Multiply both sides by $\sqrt{1-x^2}$:

$$\sqrt{1-x^2} y' = k \cosh(k \sin^{-1} x)$$

On the left side, use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$.
Let $u = \sqrt{1-x^2}$ and $v = y'$.
$\frac{du}{dx} = \frac{1}{2}(1-x^2)^{-1/2}(-2x) = \frac{-x}{\sqrt{1-x^2}}$.
$\frac{dv}{dx} = y''$.
So, the left side becomes:

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

On the right side, use the chain rule:
$\frac{d}{dx}(k \cosh(k \sin^{-1} x)) = k \cdot \sinh(k \sin^{-1} x) \cdot \frac{d}{dx}(k \sin^{-1} x)$

$$= k \cdot \sinh(k \sin^{-1} x) \cdot \frac{k}{\sqrt{1-x^2}} = \frac{k^2 \sinh(k \sin^{-1} x)}{\sqrt{1-x^2}}$$

Recall that $y = \sinh(k \sin^{-1} x)$. Substitute this into the equation:

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

$$(1-x^2) y'' - x y' - k^2 y = 0$$

This proves the given differential equation.