Question

If \(y^{1/n}+y^{-1/n}=2x\), then \((x^{2}-1)\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=\ldots\)

• A. \(n^{2}y\)
• B. \(-n^{2}y\)
• C. \(0\)
• D. \(y\)

Hint:

Square the first derivative relation to remove the radicals, then differentiate again.

Answer 1

The Correct Option is A

Step 1: Substitution
Let $y^{1/n}=e^{\theta}$. Then $2x=2\cosh\theta$, so $x=\cosh\theta$ and $y=e^{n\theta}=\left(x+\sqrt{x^{2}-1}\right)^{n}$.

Step 2: First Derivative
$y'=\dfrac{ny}{\sqrt{x^{2}-1}}$, so $\sqrt{x^{2}-1}\,y'=ny$.

Step 3: Differentiate and Multiply
Differentiating gives $\dfrac{xy'}{\sqrt{x^{2}-1}}+\sqrt{x^{2}-1}\,y''=ny'$. Multiplying by $\sqrt{x^{2}-1}$ gives $xy'+(x^{2}-1)y''=n\sqrt{x^{2}-1}\,y'=n^{2}y$.

Final Answer:
The expression reduces to n squared y. \[ \boxed{\text{(A) }n^{2}y} \]