Question

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

• A. - 23
• B. - 24
• C. 26
• D. - 26

Answer 1

Answer: (B)

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