If the solution curve f ( x , y ) = 0 of the differential equation ( 1 + log e x ) d x d y - x log e x = e y , x 0 , passes through the points (1, 0) and (2 ) , then is equal to [2023]

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Solution

Q.
If the solution curve $f (x, y) = 0$ of the differential equation $(1 + \log_{e} x) \frac{d x}{d y} - x \log_{e} x = e^{y}, x > 0,$ passes through the points (1, 0) and $(α, 2)$ , then $α^{α}$ is equal to [2023]

1 $e^{e^{2}}$

2 $e^{\sqrt{2} e^{2}}$

3 $e^{2 e^{\sqrt{2}}}$

4 $e^{2 e^{2}}$

Ans.
(4)

$The given differential equation is,$

$(1 + \log x) \frac{d x}{d y} - x \log x = e^{y}$

Put $x \log x = t \Rightarrow (x \frac{1}{x} + \log x) d x = d t$

$\Rightarrow (1 + \log x) d x = d t$

$∴$ $Given equation becomes: \frac{d t}{d y} - t = e^{y},$ which is a linear differential equation.

$∴$ $I . F . = e^{\int - d y} = e^{- y}$

So, required solution is given by

$e^{- y} \cdot t = \int e^{y} \cdot e^{- y} d y = y + C$

$\Rightarrow t = (y + C) e^{y}$

$\Rightarrow x \log x = (y + C) e^{y} \dots (i)$

Now, (i) passes through (1, 0) and $(α, 2)$

$∴ C = 0 and α \log α = 2 e^{2}$

$\Rightarrow \log α^{α} = 2 e^{2} \Rightarrow α^{α} = e^{2 e^{2}}$

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