Question

If $2x=y^{\frac{1}{5}}+y^{-\frac{1}{5}}$ and $\left(x^2-1\right)\frac{d^{2}y}{d{x}^2}+\lambda x\frac{dy}{dx}+ky=0$, then $\lambda +k$ is equal to :

• A. - 23
• B. - 24
• C. 26
• D. - 26

Answer 1

Answer: (B)

$y^{1/5}+y^{-1/5}=2x$
$\left(\frac{1}{5}{y}^{-4/5}-\frac{1}{5}{y}^{-6/5}\right)\frac{dy}{dx}=2$
$y^{\prime }\left(y^{1/5}-y^{-1/5}\right)=10y$
$y^{\prime }\left(2\sqrt{x^2-1}\right)=10y$
$y^{\prime \prime }\left(2\sqrt{x^2-1}\right)+y^{\prime }2\frac{2x}{2\sqrt{x^2-1}}=\sqrt{y^{\prime }}$
$y^{\prime \prime }\left(x^2-1\right)+x{y}^{\prime }=5\sqrt{x^2-1}\left(y^{\prime }\right)$
$y^{\prime \prime }\left(x^2-1\right)+x{y}^{\prime }-25y=0$
$\lambda =1,k=-25$