If \(y^{1/m} + y^{-1/m} = 2x\), then the value of \(\displaystyle (x^2 - 1)\frac{d^2y}{dx^2} + x\frac{dy}{dx}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(m^2y\)
Let \(u = y^{1/m}\). The given relation becomes a quadratic in u:
\(\displaystyle u + \frac{1}{u} = 2x\)
\(u^2 - 2xu + 1 = 0\)
\(\displaystyle y^{1/m} = u = x \pm \sqrt{x^2 - 1}\)
With the + sign, taking logarithms:
\(\displaystyle \frac{1}{m}\log y = \log\left(x + \sqrt{x^2 - 1}\right)\)
Differentiating with respect to x:
\(\displaystyle \frac{1}{my}\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{\sqrt{x^2 - 1} + x}{\sqrt{x^2 - 1}}\)
\(\displaystyle = \frac{1}{\sqrt{x^2 - 1}}\)
\(\displaystyle \sqrt{x^2 - 1}\,\frac{dy}{dx} = my\)
With the − sign the same steps give \(\sqrt{x^2 - 1}\,\frac{dy}{dx} = -my\). Squaring removes the sign in both cases:
\(\displaystyle (x^2 - 1)\left(\frac{dy}{dx}\right)^2 = m^2y^2\)
Differentiating again with respect to x:
\(\displaystyle 2x\left(\frac{dy}{dx}\right)^2 + 2(x^2 - 1)\frac{dy}{dx}\frac{d^2y}{dx^2} = 2m^2y\frac{dy}{dx}\)
Dividing by \(2\frac{dy}{dx}\):
\(\displaystyle (x^2 - 1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = m^2y\)