Q.If $x = \sin t$ and $y = \sin pt$, prove that $(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} + p^2 y = 0$.

Concept: Parametric Second Derivative — we express $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $t$, then substitute into the given equation.

Step 1: First derivatives.

$\frac{dx}{dt} = \cos t$, $\frac{dy}{dt} = p \cos pt$.

Hence

Step 2: Second derivative.

Differentiate the numerator:

Thus

Step 3: Substitute into $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + p^2 y$.

Since $x = \sin t$, $1 - x^2 = \cos^2 t$.

First term: …

This is a classic problem on parametric differentiation. The key idea: when both $x$ and $y$ are given in terms of a third variable (here $t$), we cannot differentiate $y$ directly with respect to $x$. Instead, we use the chain rule in two stages.

Why this works:

Since $x = \sin t$, the derivative $\frac{dx}{dt} = \cos t$. And $y = \sin pt$, so $\frac{dy}{dt} = p \cos pt$. Then by the chain rule:

This is the first derivative. For the second derivative, we differentiate $\frac{dy}{dx}$ with respect to $t$ and then divide by $\frac{dx}{dt}$ again — because $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}$.

Let's work through it step by step.

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1. Find $\frac{dy}{dx}$ Given $x = \sin t$, $y = \sin pt$.

Hence:

2. Find $\frac{d^2y}{dx^2}$ First, differentiate $\frac{dy}{dx}$ with respect to $t$:

Use the quotient rule:

Let $u = p \cos pt$, $v = \cos t$. Then $u' = -p^2 \sin pt$, $v' = -\sin t$.

Simplify numerator:

Now, $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\cos t}$ (since $\frac{dx}{dt} = \cos t$). So:

3. Substitute into the given expression We need to verify:

Recall $x = \sin t$, so $1 - x^2 = 1 - \sin^2 t = \cos^2 t$. Also $y = \sin pt$.

Compute each term:

• First term: $(1 - x^2)\frac{d^2y}{dx^2} = \cos^2 t \cdot \frac{-p^2 \sin pt \cos t + p \cos pt \sin t}{\cos^3 t}$ Simplify: $\frac{\cos^2 t}{\cos^3 t} = \frac{1}{\cos t}$, so:

 Which is:

• Second term: …

Method: Second Derivative of Parametric Functions, Used to Verify an ODE

Use this method whenever $x$ and $y$ are both given in terms of a parameter $t$, and the question asks you to prove that $x$, $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ (all expressed via $t$) satisfy some given equation.

Steps

Step 1: Differentiate $x$ and $y$ separately with respect to $t$

Find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$.

Step 2: Form the first derivative

Step 3: Find the second derivative — never divide the second $t$-derivatives directly

$\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$ …

Common Mistakes

Mistake 1: Using the wrong formula $\dfrac{d^2y}{dx^2}=\dfrac{d^2y/dt^2}{d^2x/dt^2}$

Why it's wrong: This looks like a natural extension of $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$, but it is not how the second derivative works for parametric curves — it gives a completely different, wrong expression. Correct approach: first form $\dfrac{dy}{dx}$ as a function of $t$, differentiate that with respect to $t$, then divide the result by $\dfrac{dx}{dt}$.

Mistake 2: Sign error differentiating $p\cos pt$

Why it's wrong: $\dfrac{d}{dt}(p\cos pt) = -p^2\sin pt$ — both the chain-rule factor of $p$ and the sign flip from differentiating cosine are easy to lose, especially when $p$ already appears as a coefficient outside. Correct approach: treat $p\cos pt$ as $p\cdot\cos(pt)$, apply the chain rule to $\cos(pt)$ alone, then multiply the whole result by the outer $p$.

Mistake 3: Losing track of terms when substituting into $(1-x^2)y''-xy'+p^2y$ …