Question

If y = xⁿ⁻¹ log x then x² y₂ + (3 − 2n) xy₁ is equal to

(a) −(n − 1)² y
(b) (n − 1)²y
(c) −n²y
(d) n²y

Answer 1

(a) −(n − 1)² y

Here,

$$\begin{gathered} y=x^{n-1}\log x \\ \Rightarrow y_1=\left(n-1\right)x^{n-2}\log x+\frac{x^{n-1}}{x} \\ \Rightarrow y_1=\frac{\left(n-1\right)x^{n-1}\log x+x^{n-1}}{x} \\ \Rightarrow x{y}_1=\left(n-1\right)y+x^{n-1} \\ \Rightarrow x{y}_2+y_1=\left(n-1\right)y_1+\left(n-1\right)x^{n-2} \\ \Rightarrow x{y}_2+y_1=\left(n-1\right)y_1+\frac{\left(n-1\right)x^{n-1}}{x} \\ \Rightarrow x^2{y}_2+x{y}_1=x\left(n-1\right)y_1+\left(n-1\right)x^{n-1} \\ \Rightarrow x^2{y}_2+x{y}_1=x\left(n-1\right)y_1+\left(n-1\right)\left\{x{y}_1-\left(n-1\right)y\right\} \\ \Rightarrow x^2{y}_2+x{y}_1=x\left(n-1\right)y_1+\left(n-1\right)x{y}_1-{\left(n-1\right)}^{2}y \\ \Rightarrow x^2{y}_2+x{y}_1=2x\left(n-1\right)y_1-{\left(n-1\right)}^{2}y \\ \Rightarrow x^2{y}_2+x{y}_1-2x\left(n-1\right)y_1=-{\left(n-1\right)}^{2}y \\ \Rightarrow x^2{y}_2+x{y}_1\left(1-2n+2\right)=-{\left(n-1\right)}^{2}y \\ \Rightarrow x^2{y}_2+\left(3-2n\right)x{y}_1=-{\left(n-1\right)}^{2}y \end{gathered}$$