Let x=x(y) be the solution of the differential equation 2(y+2) _e(y+2) d x+(x+4-2 _e(y+2))…

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 :

Accepted Answer

$$
\begin{aligned}
& 2(y+2) \ln (y+2) d x+(x+4-2 \ln (y+2)) d y=0 \\
& 2 \ln (y+2)+(x+4-2 \ln (y+2)) \frac{1}{y+2} \cdot \frac{d y}{d x}=0 \\
& \text { let, } \ln (y+2)=t \\
& \frac{1}{y+2} \cdot \frac{d y}{d x}=\frac{d t}{d x} \\
& 2 t+(x+4-2 t) \cdot \frac{d t}{d x}=0 \\
& (x+4-2 t) \frac{d t}{d x}=-2 t \\
& \frac{d x}{d t}=\frac{2 t-4-x}{2 t} \\
& \frac{d x}{d t}+\frac{x}{2 t}=\frac{2 t-4}{2 t}
\end{aligned}
$$

$$
\begin{aligned}
& x \cdot t^{1 / 2}=\int \frac{2 t-4}{2 t} \cdot t^{1 / 2} \cdot d t \\
& x \cdot t^{1 / 2}=\int\left(t^{1 / 2}-\frac{2}{t^{1 / 2}}\right) \cdot d t \\
& =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-2 \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\
& x \cdot t^{\frac{1}{2}}=\frac{2 t^{\frac{3}{2}}}{3}-4 \mathrm{t}^{\frac{1}{2}}+C \\
& x=\frac{2}{3} \cdot \mathrm{t}-4+\mathrm{C} \cdot \mathrm{t}^{\frac{-1}{2}} \\
& x=\frac{2}{3} \ln (\mathrm{y}+2)-4+C \cdot(\ln (\mathrm{y}+2))^{\frac{-1}{2}}
\end{aligned}
$$

Put $y=e^4-2, x=1$

$$
\begin{aligned}
& 1=\frac{2}{3} \times 4-4+C \times \frac{1}{2} \\
& \frac{C}{2}=5-\frac{8}{3}=\frac{7}{3} \\
& C=\frac{14}{3} \\
& x=\frac{2}{3} \times 9-4+\frac{14}{3} \times \frac{1}{3} \\
& =2+\frac{14}{9} \\
& =\frac{32}{9}
\end{aligned}
$$

Solutions · 1