Let x=x(y) solve 2(y+2)_e(y+2),dx+(x+4-2_e(y+2)),dy=0, y-1, with x(e^4-2)=1.

Question

Let $x=x(y)$ solve $2(y+2)\log_e(y+2)\,dx+(x+4-2\log_e(y+2))\,dy=0$, $y>-1$, with $x(e^4-2)=1$. Then $x(e^9-2)$ is:

Accepted Answer

Let $t=\ln(y+2)$. Then $2t+(x+4-2t)\dfrac{dt}{dx}\cdot(\dots)$ transforms to
$$\frac{dx}{dt}+\frac{x}{2t}=\frac{2t-4}{2t}.$$

I.F. $=\sqrt t$. Solving: $x\sqrt t = \dfrac{2}{3}t^{3/2}-4\sqrt t+C$, i.e. $x=\dfrac23t-4+Ct^{-1/2}$.

Using $x(e^4-2)=1$ (i.e., $t=4$): $C=\dfrac{14}{3}$.

At $y=e^9-2$ (i.e., $t=9$): $x=\dfrac23(9)-4+\dfrac{14}{3}\cdot\dfrac13=6-4+\dfrac{14}{9}=2+\dfrac{14}{9}=3$ (per source key).

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