Question

If $y=y\left(x\right)$ is the solution of the differential equation $\frac{5+e^x}{2+y}\cdot \frac{dy}{dx}+e^x=0$ satisfying $y\left(0\right)=1$, then a value of $y\left({\log }_{e}13\right)$ is:

• A. $1$
• B. $0$
• C. $2$
• D. $-1$

Answer 1

The correct option is D $-1$

$\frac{5+e^x}{2+y}\cdot \frac{dy}{dx}=-e^x$
$\Rightarrow \int \frac{dy}{2+y}=\int \frac{-e^x}{e^x+5}dx$
$\Rightarrow \ln (y+2)=-\ln (e^x+5)+C$
$\Rightarrow (y+2)(e^x+5)=C$

$\because y\left(0\right)=1$
$\Rightarrow (1+2)(e^0+5)=C$
$\Rightarrow C=18$
$\Rightarrow y+2=\frac{18}{e^x+5}$
At $x=\ln 13,$
$y+2=\frac{18}{13+5}=1$
$\therefore y=-1$