Let y = y ( x ) be the solution of the differential equation x log e x ? d y d x + y = x 2 log e x , ( x 1 ) . If y ( 2 ) = 2 , then y ( e ) is equal to [2023]

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Solution

Q.
Let $y = y (x)$ be the solution of the differential equation $x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x, (x > 1) .$ If $y (2) = 2$ , then $y (e)$ is equal to [2023]

1 $\frac{1 + e^{2}}{2}$

2 $\frac{4 + e^{2}}{4}$

3 $\frac{1 + e^{2}}{4}$

4 $\frac{2 + e^{2}}{2}$

Ans.
(2)

$Given: x \log_{e} x \frac{d y}{d x} + y = x^{2} \log_{e} x (x > 1)$

$\Rightarrow \frac{d y}{d x} + \frac{y}{x \ln x} = x$ which is a linear differential equation.

whose Integrating factor is given by

$I.F. = e^{\int \frac{1}{x \ln x} d x} =$ $\ln$ $| x |$

$y \ln$ $| x |$ $= \frac{x^{2}}{2} \ln x - \int \frac{1}{x} \cdot \frac{x^{2}}{2} d x$ ...(i)

$\Rightarrow y \ln$ $| x |$ $= \ln$ $| x |$ $\frac{x^{2}}{2} - \frac{x^{2}}{4} + C$

$Also, y (2) = 2 \Rightarrow C = 1$

On substituting the value of $C$ in (i), we get

$y (x) = \frac{x^{2}}{2} - \frac{x^{2}}{4 | \ln (x) |} + \frac{1}{| \ln (x) |}$

$\Rightarrow y (e) = \frac{e^{2}}{2} - \frac{e^{2}}{4} + 1 = 1 + \frac{e^{2}}{4}$ or $\frac{4 + e^{2}}{4}$

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