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Let $y=y(x)$ be the solution of the differential equation, $x\frac{dy}{dx}+y=x\log_{e}x,\: \: \left ( x> 1 \right ).$ If $2y(2)=\log_{e}4-1,$ then $y(e)$ is equal to :
Option 1)
$\frac{e^{2}}{4}$

Option 2)
$-\frac{e}{2}$
Option 3)
$-\frac{e^{2}}{2}$
Option 4)
$\frac{e}{4}$

Answers (1)

best_answer

Linear Differential Equation -

$\frac{dy}{dx}+Py= Q$

- wherein

P, Q are functions of x alone.

Linear Differential Equation -

Multiply by $e^{SPdx}$ which is the Integrating factor

- wherein

P is the function of x alone

$x\frac{dy}{dx}+y=x\log_{e}x,\: \: \left ( x> 1 \right ).$

$\frac{dy}{dx}+\frac{y}{x}=\log_{e}x$

$I.F. = e^{\int \frac{dx}{x}}=x$

$yx=\int x\log_{e}x\: \: dx$

$=>xy=(\log_{e}x)\times \frac{x^{2}}{2}-\frac{x^{2}}{4}+C$

Putting x = 2 we get C = 0

$y=\frac{x}{2}\log_{e}x-\frac{x}{4}$

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

Option 1)

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

Option 2)

$-\frac{e}{2}$

Option 3)

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

Option 4)
$\frac{e}{4}$

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