Let y = y(x) be the solution of the differential equation ( 2 x log e x ) d y d x + 2 y = 3 x log e x , x > 0 and y ( e - 1 ) = 0 . Then y(e) is equal to [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $(2 x \log_{e} x) \frac{d y}{d x} + 2 y = \frac{3}{x} \log_{e} x$ , x > 0 and $y (e^{- 1}) = 0$ . Then y(e) is equal to [2024]

1 $- \frac{3}{2 e}$

2 $- \frac{2}{e}$

3 $- \frac{3}{e}$

4 $- \frac{2}{3 e}$

Ans.
(3)

We Have, $\frac{d y}{d x} + \frac{2 y}{2 x \log_{e} x} = \frac{3 \log_{e} x}{x (2 x \log_{e} x)}$

$\Rightarrow \frac{d y}{d x} + \frac{y}{x \log_{e} x} = \frac{3}{2 x^{2}}$

$IF = e^{\int \frac{1}{x \log_{e} x} d x} = e^{\log (\log_{e} x)} = \log_{e} x$

$\Rightarrow y \log_{e} x = \int \frac{3}{2 x^{2}} \log_{e} x d x$

$= \frac{3}{2} \log_{e} x \int x^{- 2} d x - \int (\frac{3}{2 x} \int x^{- 2} d x) d x$

$= \frac{3}{2} \log_{e} x (\frac{- 1}{x}) + \int \frac{3}{2 x^{2}} d x + c = \frac{- 3 \log_{e} x}{2 x} + (\frac{- 3}{2 x}) + c$

Now, $y (e^{- 1}) = 0$

$\Rightarrow 0 = \frac{3 e}{2} - \frac{3 e}{2} + c \Rightarrow c = 0$

$∴ y \log_{e} x = \frac{- 3 \log_{e} x}{2 x} - \frac{3}{2 x} \Rightarrow y = \frac{- 3}{2 x} - \frac{3}{2 x \log_{e} x}$

$\Rightarrow y (e) = \frac{- 3}{2 e} - \frac{3}{2 e} = \frac{- 3}{e}$

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