Question

If $2x=y^{1/5}+y^{-1/5}$ and $(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=ky,$ then the value of $k$ is

Answer 1

$2x=y^{1/5}+y^{-1/5}$
On differentiating with respect to $x$
$\frac{1}{5}\cdot y^{1/5-1}\cdot y_1-\frac{1}{5}\cdot y^{-1/5-1}\cdot y_1=2\Rightarrow \frac{1}{5}\cdot \frac{y^{1/5}\cdot y_1}{y}-\frac{1}{5}\cdot \frac{y^{-1/5}\cdot y_1}{y}=2\Rightarrow y^{1/5}-y^{-1/5}=\frac{10y}{y_1}\Rightarrow {(y^{1/5}-y^{-1/5})}^2=\frac{100y^2}{y_1^2}\Rightarrow {(y^{1/5}+y^{-1/5})}^2-4=\frac{100y^2}{y_1^2}\Rightarrow 4(x^2-1)=\frac{100y^2}{y_1^2}\Rightarrow y_1^2(x^2-1)=25y^2$
Again differentiating with respect to $x$
$(x^2-1)2y_1{y}_2+y_1^{2}2x=25\cdot 2\cdot y{y}_1\Rightarrow (x^2-1)y_2+x{y}_1=25y$
or $(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=25y$
$\therefore k=25$