If $y(y(x)$ is the solution of the differential equation $\frac{5+e^x}{2+y}\frac{dy}{dx}+e^x=0$ satisfying $y(0) = 1, $ then a value of $y (log_e13)$ is
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : -1
The given differential equation is
$\frac{5+e^x}{2+y}\frac{dy}{dx}+e^x=0$
$⇒\frac{1}{2+y}dy+\frac{e^x}{5+e^x}dx=0$
Integrating, we obtain
$log_e(2+y) +log_e(5+e^x) =logC $
$⇒(2+y) (5+ e^x) = C$ ............(i)
Putting $x=0, y = 1 $ in (i) , we obtain $C = 18$.
Putting $C= 18$ in (i) , we obtain
$y=\frac{18}{5+e^x}-2$
Putting $x=log_e 13 $ i.e $e^x= 13 $ in (ii), we obtain
$y = \frac{18}{5+13}-2=1$
Hence, $y \left(log_e 13\right) = -1$