Question:
If $y=y(x)$ is the solution of the differential equation $\frac{5+\mathrm{e}^{x}}{2+y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^{x}=0$ satisfying $y(0)=1$, then a value of $y\left(\log _{\mathrm{e}} 13\right)$ is :
Correct Option: , 2
Solution:
$\frac{5+e^{x}}{2+y} \cdot \frac{d y}{d x}=-e^{x}$
$\int \frac{d y}{2+y}=-\int \frac{e^{x}}{5+e^{x}} d x$
$\Rightarrow \log _{e}|2+y| \cdot \log _{e}\left|5+e^{x}\right|=\log _{e} C$
$\Rightarrow\left|(2+y)\left(5+e^{x}\right)\right|=C \quad \because y(0)=1$
$C=18$
$\therefore(2+y) \cdot\left(5+e^{x}\right)=18$
When $x=\log _{e} 13$ then $(2+y) \cdot 18=18$
$\Rightarrow 2+y=\pm 1$
$\therefore y=-1,-3$
$\therefore y(\ln 13)=-1$