If `x=sint` and `y=sinp t ,` prove that `(1-x^2)(d^2y)/(dx^2)-x(dy)/(dx)+p^2y=0.`

Explanation

To prove the equation \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0\) given \(x=\sin t\) and \(y=\sin(pt)\), we will follow these steps:

### Step 1: Find \(\frac{dy}{dx}\)

Since both \(y\) and \(x\) are functions of \(t\), we can use the chain rule to find \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
...