JEE PYQ: Differential Equation Question 23
Question 23 - 2020 (05 Sep Shift 1)
If $y = y(x)$ is the solution of the differential equation $\frac{5+e^x}{2+y}\cdot\frac{dy}{dx} + e^x = 0$ satisfying $y(0) = 1$, then a value of $y(\log_e 13)$ is:
(1) 1
(2) $-1$
(3) 0
(4) 2
Show Answer
Answer: (1)
Solution
$\frac{dy}{2+y} = -\frac{e^x}{5+e^x},dx$. Integrating: $\ln|2+y| = -\ln|5+e^x| + \ln C$. So $(2+y)(5+e^x) = C$. At $(0,1)$: $C = 18$. At $x = \ln 13$: $(2+y)(5+13) = 18$, so $2+y = 1$, $y = -1$.
BETA
