Q.If $y = (\sec^{-1} x)^2$, $x > 0$, show that $x^2(x^2 - 1)\dfrac{d^2y}{dx^2} + (2x^3 - x)\dfrac{dy}{dx} - 2 = 0$.

The key idea is to use implicit differentiation after simplifying the first derivative by clearing denominators and squaring.

1. Given $y = (\sec^{-1} x)^2$. Differentiating with respect to $x$:

   $\frac{dy}{dx} = 2(\sec^{-1} x) \cdot \frac{1}{x\sqrt{x^2 - 1}}$

   $\frac{dy}{dx} = \frac{2\sec^{-1} x}{x\sqrt{x^2 - 1}}$

2. Rearrange the first derivative to simplify subsequent differentiation. Multiply by the denominator and square both sides:

   $x\sqrt{x^2 - 1} \frac{dy}{dx} = 2\sec^{-1} x$

   $x^2(x^2 - 1) \left(\frac{dy}{dx}\right)^2 = (2\sec^{-1} x)^2$

   Since $y = (\sec^{-1} x)^2$, we substitute $4y$ for $(2\sec^{-1} x)^2$:

   $x^2(x^2 - 1) \left(\frac{dy}{dx}\right)^2 = 4y$

3. Differentiate this equation implicitly with respect to $x$:

   $\frac{d}{dx}\left[x^2(x^2 - 1) \left(\frac{dy}{dx}\right)^2\right] = \frac{d}{dx}[4y]$

   Using the product rule on the left side:

   $[2x(x^2 - 1) + x^2(2x)] \left(\frac{dy}{dx}\right)^2 + x^2(x^2 - 1) \cdot 2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} = 4\frac{dy}{dx}$

   $(2x^3 - 2x + 2x^3) \left(\frac{dy}{dx}\right)^2 + 2x^2(x^2 - 1)\frac{dy}{dx}\frac{d^2y}{dx^2} = 4\frac{dy}{dx}$ …

When asked to show that a given function satisfies a differential equation, the general strategy is to find the required derivatives of the function and substitute them into the equation. If the equation holds true (i.e., both sides are equal), then the function satisfies it.

For this problem, we need to find the first derivative, $\frac{dy}{dx}$, and the second derivative, $\frac{d^2y}{dx^2}$, of $y = (\sec^{-1} x)^2$. A common pitfall in such problems is to directly compute the second derivative from a complex first derivative expression using the quotient rule, which can become very cumbersome. A more elegant approach often involves rearranging the first derivative equation to eliminate inverse trigonometric terms or clear denominators before differentiating a second time. This is known as implicit differentiation and often simplifies the algebra significantly.

Let's proceed step-by-step.

1. Calculate the first derivative, $\frac{dy}{dx}$:

   We are given $y = (\sec^{-1} x)^2$.

   To differentiate this, we use the chain rule: $\frac{d}{dx} (u^n) = n u^{n-1} \frac{du}{dx}$. Here, $u = \sec^{-1} x$ and $n = 2$.

   We also need the derivative of $\sec^{-1} x$.

   $\frac{d}{dx} (\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2 - 1}}$

   Since the problem states $x > 0$, we can simplify this to $\frac{1}{x\sqrt{x^2 - 1}}$.

   Applying the chain rule:

So, the first derivative is:

2. Rearrange the first derivative for easier second differentiation: Directly differentiating equation (1) using the quotient rule would be complex. Notice that the term $\sec^{-1} x$ is still present. We can isolate it and then differentiate. From equation (1), multiply both sides by $x\sqrt{x^2 - 1}$:

This form is much easier to differentiate implicitly.

3. Calculate the second derivative, $\frac{d^2y}{dx^2}$:

Now, differentiate both sides of equation (2) with respect to $x$.

The left side requires the product rule: $\frac{d}{dx}(uvw) = u'vw + uv'w + uvw'$. Here, we can treat it as $\frac{d}{dx} \left( (x\sqrt{x^2 - 1}) \cdot \frac{dy}{dx} \right)$.

Let $u = x\sqrt{x^2 - 1}$ and $v = \frac{dy}{dx}$.

Then $\frac{d}{dx}(uv) = u'v + uv'$.

First, let's find $u' = \frac{d}{dx}(x\sqrt{x^2 - 1})$. This also requires the product rule:

$$ = \sqrt{x^2 - 1} + x \cdot \frac{1}{2\sqrt{x^2 - 1}} \cdot (2x) $$ …