Question

If \(x = sin(t),y = cos(pt)\), then \((1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx} =\)

• A. \(0\)
• B. \(-p^2y\)
• C. \(p^2y\)
• D. \(p\)

Hint:

Express dy/dx and d2y/dx2 using t, then substitute into the given expression.

Answer 1

The Correct Option is B

Step 1: First derivative:
\(\dfrac{dx}{dt}=\cos t\), \(\dfrac{dy}{dt}=-p\sin(pt)\). So \(\dfrac{dy}{dx}=\dfrac{-p\sin(pt)}{\cos t}\).

Step 2: Rearrange:
\(\sqrt{1-x^2}\,\dfrac{dy}{dx}=-p\sin(pt)\), since \(\cos t=\sqrt{1-x^2}\).

Step 3: Differentiate again with respect to x:
\(\sqrt{1-x^2}\,y''-\dfrac{x}{\sqrt{1-x^2}}\,y'=-p^2\cos(pt)\cdot\dfrac{dt}{dx}=\dfrac{-p^2y}{\sqrt{1-x^2}}\) after using \(dt/dx=1/\sqrt{1-x^2}\).

Step 4: Multiply through by sqrt(1-x^2):
\[ (1-x^2)y''-xy'=-p^2y \]
Option B.

Final Answer:
The expression equals -p^2 y. \[ \boxed{\text{(B) }-p^2y} \]