Let $x=x(y)$ be the solution of the differential equation

$2(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0, y>-1$

with $x\left(e^{4}-2\right)=1$. Then $x\left(e^{9}-2\right)$ is equal to :

$$ \begin{aligned} & 2(y+2) \ln (y+2) d x+(x+4-2 \ln (y+2)) d y=0 \\\\ & 2 \ln (y+2)+(x+4-2 \ln (y+2)) \frac{1}{y+2} \cdot \frac{d y}{d x}=0 \\\\ & \text { let, } \ln (y+2)=t \\\\ & \frac{1}{y+2} \cdot \frac{d y}{d x}=\frac{d t}{d x} \\\\ & 2 t+(x+4-2 t) \cdot \frac{d t}{d x}=0 \\\\ & (x+4-2 t) \frac{d t}{d x}=-2 t \\\\ & \frac{d x}{d t}=\frac{2 t-4-x}{2 t} \\\\ & \frac{d x}{d t}+\frac{x}{2 t}=\frac{2 t-4}{2 t} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & x \cdot t^{1 / 2}=\int \frac{2 t-4}{2 t} \cdot t^{1 / 2} \cdot d t \\\\ & x \cdot t^{1 / 2}=\int\left(t^{1 / 2}-\frac{2}{t^{1 / 2}}\right) \cdot d t \\\\ & =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-2 \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\\\ & x \cdot t^{\frac{1}{2}}=\frac{2 t^{\frac{3}{2}}}{3}-4 \mathrm{t}^{\frac{1}{2}}+C \\\\ & x=\frac{2}{3} \cdot \mathrm{t}-4+\mathrm{C} \cdot \mathrm{t}^{\frac{-1}{2}} \\\\ & x=\frac{2}{3} \ln (\mathrm{y}+2)-4+C \cdot(\ln (\mathrm{y}+2))^{\frac{-1}{2}} \end{aligned} $$ <br/><br/>Put $y=e^4-2, x=1$ <br/><br/>$$ \begin{aligned} & 1=\frac{2}{3} \times 4-4+C \times \frac{1}{2} \\\\ & \frac{C}{2}=5-\frac{8}{3}=\frac{7}{3} \\\\ & C=\frac{14}{3} \\\\ & x=\frac{2}{3} \times 9-4+\frac{14}{3} \times \frac{1}{3} \\\\ & =2+\frac{14}{9} \\\\ & =\frac{32}{9} \end{aligned} $$