Let x = x (y) be the solution of the differential equation 2(y + 2)...

Given,

$2\left(y + 2\right)\log_{e}\left(y + 2\right)dx + \left(x + 4 - 2\log_{e}\left(y + 2\right)\right)dy = 0$

Let $x + 4 = u, y + 2 = v$

$dx = du, dy = dv$

So, the equation becomes,

\left(2vlnv\right)du = - \left(u - 2\ln v\right)dv$</p><p>$⇒ 2vlnv\dfrac{du}{dv} + u = 2\ln v$</p><p>$⇒ \dfrac{du}{dv} + \dfrac{1}{2vlnv} . u = \dfrac{1}{v}$</p><p>Which is a linear differential equation,</p><p>So,$IF = {e}^{\dfrac{1}{2} ∫ \dfrac{1}{vlnv}} = {e}^{\dfrac{1}{2}\ln \left(\ln v\right)} = \left(\ln v\right)\dfrac{1}{2}$</p><p>Now solution of differential equation is given by,</p><p>$u \cdot \left(\ln v\right)\dfrac{1}{2} = ∫ \dfrac{1}{v} \cdot \left(\ln v\right)\dfrac{1}{2}dv$</p><p>$⇒ u \cdot \left(\ln v\right)\dfrac{1}{2} = \dfrac{2}{3}\left(\ln v\right)\dfrac{3}{2} + c . . . . . . . \left(i\right)$</p><p>Now using given value,$y = {e}^{4} - 2 ⇒ x = 1$</p><p>$∴ v = {e}^{4} ⇒ u = 5$</p><p>$5 \cdot \left(4\dfrac{1}{2}\right) = \dfrac{2}{3} \cdot \left(4\right)\dfrac{3}{2} + c$</p><p>$⇒ 10 = \dfrac{16}{3} + c$</p><p>$⇒ c = \dfrac{14}{3}$</p><p>Now finding,$y = {e}^{9} - 2 ⇒ v = y + 2 = {e}^{9}$</p><p>Now putting the value in equation \left(i\right)$ we get,

$⇒ u \cdot 3 = \dfrac{2}{3} \times 27 + \dfrac{14}{3} = 18 + \dfrac{14}{3}$

$⇒ x + 4 = u = 6 + \dfrac{14}{9}$

$⇒ x = 2 + \dfrac{14}{9} = \dfrac{32}{9}$