Let y=y(x) be the solution of the differential equation,

Question:

Let $y=y(x)$ be the solution of the differential equation, $x \frac{d y}{d x}+y=x \log _{\mathrm{e}} \mathrm{x},(\mathrm{x}>1) .$ If $2 \mathrm{y}(2)=\log _{\mathrm{e}} 4-1$, then $\mathrm{y}(\mathrm{e})$ is equal to:

  1. (1) $-\frac{e}{2}$

  2. (2) $-\frac{e^{2}}{2}$

  3. (3) $\frac{e}{4}$

  4. (4) $\frac{e^{2}}{4}$


Correct Option: , 3

Solution:

Consider the differential equation,

$\frac{d y}{d x}+\frac{y}{x}=\log _{e} x$

$\because \quad I F=e^{\int \frac{1}{x} d x}=x$

$\therefore \quad y x=\int x \ln x d x$

$\Rightarrow \quad x y=\ln x \cdot \frac{x^{2}}{2}-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x$

$\Rightarrow \quad x y=\frac{x^{2}}{2} \cdot \ln x-\frac{x^{2}}{4}+c$

Given, $2 y(2)=\log _{e} 4-1$

$\therefore \quad 2 y=2 \ln 2-1+c$

$\Rightarrow \quad \ln 4-1=\ln 4-1+c$

i.e. $c=0$

$\Rightarrow \quad x y=\frac{x^{2}}{2} \ln x-\frac{x^{2}}{4}$

$\Rightarrow \quad y=\frac{x}{2} \ln x-\frac{x}{4}$

$\Rightarrow y(e)=\frac{e}{2}-\frac{e}{4}=\frac{e}{4}$

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