Q106 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

If $y=e^x(\log x)$, then $x y_2+(x-1) y=$

  • Correct answer
    A
    $(2 x-1) y_1$
  • B
    $(x-1) y_1$
  • C
    $(4-2 x) y_1$
  • D
    $(3 x-1) y_1$

Solution

Given, $y=e^x(\log x)$ $$ \frac{d y}{d x}=\frac{e^x}{x}+e^x \ln x $$ $$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{e^x}{x}+y \Rightarrow x \frac{d^2 y}{d x^2}=x \frac{x \cdot e^x-e^x}{x^2}+\frac{d y}{d x} x \\ & \Rightarrow x \frac{d^2 y}{d x^2}=\frac{x^2 e^x-x e^x}{x^2}+x \frac{d y}{d x} \\ & \Rightarrow x \frac{d^2 y}{d x^2}+(x-1) y=\frac{x e^x\{x-1\}}{x^2}+x \frac{d y}{d x}+(x-1) y \\ & \Rightarrow \frac{x d^2 y}{d x^2}+(x-1) y=\frac{e^x(x-1)}{x}+(x-1) y+x \frac{d y}{d x} \\ & =e^x-\frac{e^x}{x}+(x-1) e^x \ln x+x \frac{d y}{d x} \\ & =(x-1) e^x\left\{\frac{1}{x}+\ln x\right\}+x \frac{d y}{d x} \\ & \end{aligned} $$ $$ \begin{aligned} & \Rightarrow \quad x \frac{d^2 y}{d x^2}+(x-1) y=(x-1) \quad\left[\frac{e^x}{x}+e^x \ln x\right]+x \frac{d y}{d x} \\ & \Rightarrow x \frac{d^2 y}{d x^2}+(x-1) y=(x-1+x) \frac{d y}{d x} \\ & x \frac{d^2 y}{d x^2}+(x-1) y=(2 x-1) \frac{d y}{d x} \\ & \end{aligned} $$ [by Eq. (i)]

Hence, the correct answer is option: 'A'.

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