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 :

Question

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 :

$\frac{4}{9}$