If the solution curve of the differential equation ( y - 2 _ e x )...

Given,

$( y - 2\log_{e}x ) dx + ( x{\log}_{e}{x}^{2} ) dy = 0, x > 1$

$⇒ 2xlnx\dfrac{dy}{dx} + y = 2\ln x\left\{\text{where}\log_{e}x = \ln x\right\}$

$⇒ \dfrac{dy}{dx} + \dfrac{y}{2xlnx} = \dfrac{1}{x}$

Which is linear differential equation,

So, $I . F = {e}^{∫ \dfrac{1}{2xlnx}} = \sqrt{\ln x}$

Now, solution of the equation is given by,

$y \cdot \sqrt{\ln x} = ∫ \dfrac{\sqrt{\ln x}}{x}dx$

$⇒ y \cdot \sqrt{\ln x} = \dfrac{2}{3}\left(\ln x\right)\dfrac{3}{2} + C . . . ( i )$

Given, $eq . ( i )$ passes through point \left(e, \dfrac{4}{3}\right)$</p><p>So,$C = \dfrac{2}{3}$</p><p>Hence, solution of differential equation will be,</p><p>$y\sqrt{\ln x} = \dfrac{2}{3}\left(\ln x\right)\dfrac{3}{2} + \dfrac{2}{3}$</p><p>Also given, above equation passes through point$( {e}^{4}, \alpha )$</p><p>$\alpha\sqrt{ln{e}^{4}} = \dfrac{2}{3}\left(ln{e}^{4}\right)\dfrac{3}{2} + \dfrac{2}{3}$</p><p>$⇒ 2\alpha = \dfrac{2}{3}\left(4\right)\dfrac{3}{2} + \dfrac{2}{3}$</p><p>$⇒ \alpha = \dfrac{1}{3} \times 8 + \dfrac{1}{3}$</p><p>$⇒ \alpha = 3