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(C) The given differential equation is $2(y + 2) \ln(y + 2) dx + (x + 4 - 2 \ln(y + 2)) dy = 0$.Rearranging,we get $\frac{dx}{dy} = -\frac{x + 4 - 2 \

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$\frac{32}{9}$

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(C) The given differential equation is $2(y + 2) \ln(y + 2) dx + (x + 4 - 2 \ln(y + 2)) dy = 0$.Rearranging,we get $\frac{dx}{dy} = -\frac{x + 4 - 2 \ln(y + 2)}{2(y + 2) \ln(y + 2)}$.This can be written as $2(y + 2) \ln(y + 2) \frac{dx}{dy} + x = 2 \ln(y + 2) - 4$.Let $t = \ln(y + 2)$,then $dt = \frac{1}{y + 2} dy$,so $dy = (y + 2) dt$.The equation becomes $2t \frac{dx}{dt} + x = 2t - 4$,or $\frac{dx}{dt} + \frac{x}{2t} = 1 - \frac{2}{t}$.This is a linear differential equation in $x$ with respect to $t$.The integrating factor is $IF = e^{\int \frac{1}{2t} dt} = e^{\frac{1}{2} \ln t} = \sqrt{t}$.Multiplying by $IF$,we get $\frac{d}{dt}(x \sqrt{t}) = \sqrt{t} - \frac{2}{\sqrt{t}}$.Integrating both sides,$x \sqrt{t} = \int (t^{1/2} - 2t^{-1/2}) dt = \frac{2}{3} t^{3/2} - 4t^{1/2} + C$.Dividing by $\sqrt{t}$,$x = \frac{2}{3} t - 4 + \frac{C}{\sqrt{t}}$.Substituting $t = \ln(y + 2)$,$x = \frac{2}{3} \ln(y + 2) - 4 + \frac{C}{\sqrt{\ln(y + 2)}}$.Given $x(e^4 - 2) = 1$,so $y = e^4 - 2$,$t = \ln(e^4) = 4$,$x = 1$.$1 = \frac{2}{3}(4) - 4 + \frac{C}{\sqrt{4}} \implies 1 = \frac{8}{3} - 4 + \frac{C}{2} \implies 1 = -\frac{4}{3} + \frac{C}{2} \implies \frac{C}{2} = \frac{7}{3} \implies C = \frac{14}{3}$.Now find $x(e^9 - 2)$,so $y = e^9 - 2$,$t = \ln(e^9) = 9$.$x = \frac{2}{3}(9) - 4 + \frac{14/3}{\sqrt{9}} = 6 - 4 + \frac{14/3}{3} = 2 + \frac{14}{9} = \frac{32}{9}$.

Let $x = x(y)$ be the solution of the differential equation $2(y + 2) \log_e(y + 2) dx + (x + 4 - 2 \log_e(y + 2)) dy = 0$,$y > -1$ with $x(e^4 - 2) = 1$. Then $x(e^9 - 2)$ is equal to

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