Question

If $y^{\frac{1}{n}}+y^{-\frac{1}{n}}=2x,\text{then find}\left(x^2-1\right)y_2+x{y}_1=$

Answer 1

$$\begin{gathered} \left(c\right)n^{2}y \\ {y}^{\frac{1}{n}}+y^{-\frac{1}{n}}=2x \\ \text{Differentiating the above equation with respect to}\mathit{\text{x}} \\ \left(\frac{1}{n}{y}^{\frac{\mathit{1}}{n}\mathit{-}\mathit{1}}-\frac{1}{n}{y}^{-\frac{1}{n}-1}\right)y_1\mathit{=}2 \\ \frac{1}{ny}\left(y^{\frac{\mathit{1}}{n}}-y^{-\frac{1}{n}}\right)y_1\mathit{=}2 \\ \left(y^{\frac{\mathit{1}}{n}}-y^{-\frac{1}{n}}\right)y_1\mathit{=}2ny.....\left(1\right) \\ \left(y^{\frac{\mathit{1}}{n}}-y^{-\frac{1}{n}}\right)y_2+y_1\left(\frac{1}{n}{y}^{\frac{1}{n}\mathit{-}1}+\frac{1}{n}{y}^{-\frac{1}{n}-1}\right)y_1=2n{y}_1 \\ ny\left(y^{\frac{\mathit{1}}{n}}-y^{-\frac{1}{n}}\right)y_2+{y_1}^2\left(y^{\frac{1}{n}}+y^{-\frac{1}{n}}\right)=2n^{2}y{y}_1 \\ \text{Dividing the above equation by}{\mathit{\text{y}}}_{\mathit{1}} \\ \frac{ny}{y_1}\left(y^{\frac{\mathit{1}}{n}}-y^{-\frac{1}{n}}\right)y_2+y_1\left(y^{\frac{1}{n}}+y^{-\frac{1}{n}}\right)=2n^{2}y \\ \text{Putting}{\mathit{\text{y}}}_1\text{from equation}\left(1\right) \\ \frac{{\left(y^{\frac{\mathit{1}}{n}}-y^{-\frac{1}{n}}\right)}^2}{2}{y}_2+y_1\left(y^{\frac{1}{n}}+y^{-\frac{1}{n}}\right)=2n^{2}y.....\left(2\right) \\ \mathrm{Now}, \\ {\left(y^{\frac{1}{n}}-y^{-\frac{1}{n}}\right)}^2={\left(y^{\frac{1}{n}}+y^{-\frac{1}{n}}\right)}^2-4 \\ {\left(y^{\frac{1}{n}}-y^{-\frac{1}{n}}\right)}^2=4x^2-4.....\left(3\right) \\ \text{Putting the value of}\left(3\right)\text{in}\left(2\right) \\ \frac{4\left(x^2-1\right)y_2}{2}+2x{y}_1=2n^{2}y \\ \left(x^2-1\right)y_2+x{y}_1=n^{2}y \end{gathered}$$