Q.If $y = e^{a \cos^{-1} x}$, $-1 \leq x \leq 1$, show that $(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0$.

Idea: differentiate $y=e^{a\cos^{-1}x}$ once to relate $y'$ to $y$, then differentiate again and substitute.

First derivative (chain rule, $\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}$):

So $\sqrt{1-x^2}\,\dfrac{dy}{dx} = -a\,y$. Square: $(1-x^2)\left(\dfrac{dy}{dx}\right)^2 = a^2 y^2$.

Differentiate this w.r.t. $x$: …

Because $y$ is an exponential of $\cos^{-1}x$, its first derivative comes out proportional to $y$ itself. Squaring removes the square root cleanly, and one more differentiation produces the required second-order relation.

Step 1 — first derivative

With $u=a\cos^{-1}x$ and $y=e^{u}$, the chain rule gives

Rearrange to clear the root:

Step 2 — square to remove the root

Step 3 — differentiate both sides with respect to $x$

Left side (product rule on $(1-x^2)$ and $(y')^2$):

Right side:

So

Step 4 — cancel the common factor …

Method: Proving an ODE for $y=e^{a\cdot g(x)}$ by Squaring to Remove a Root

When the first derivative of an exponential-of-inverse-trig function comes out proportional to $y$ but divided by a square root, squaring both sides before differentiating again avoids having to differentiate a square root directly.

Steps

Step 1: Differentiate once using the chain rule

For $y=e^{a\cos^{-1}x}$: $y_1 = y\cdot a\cdot\left(-\frac{1}{\sqrt{1-x^2}}\right)$, i.e. $\sqrt{1-x^2}\,y_1=-ay$.

Step 2: Square both sides to eliminate the square root

$(1-x^2)y_1^2=a^2y^2$.

Step 3: Differentiate this polynomial-style equation implicitly with respect to $x$ …

Common Mistakes

Mistake 1: Attempting to differentiate the square-root form of $y_1$ directly instead of squaring first.

Why it's wrong: differentiating $-\frac{ay}{\sqrt{1-x^2}}$ directly requires the quotient rule combined with the chain rule on a square root — technically possible but far more error-prone than clearing the root first. Correct approach: square both sides of the first-derivative relation before differentiating again, exactly as with the logarithmic-argument ODE method.

Mistake 2: Sign error on $\dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^2}}$. …