JEE PYQ: Differential Equation Question 19
Question 19 - 2020 (03 Sep Shift 1)
The solution curve of the differential equation $(1+e^{-x})(1+y^2)\frac{dy}{dx} = y^2$, which passes through the point $(0, 1)$, is:
(1) $y^2 + 1 = y\left[\log_e\left(\frac{1+e^{-x}}{2}\right) + 2\right]$
(2) $y^2 + 1 = y\left[\log_e\left(\frac{1+e^x}{2}\right) + 2\right]$
(3) $y^2 = 1 + y\log_e\left(\frac{1+e^x}{2}\right)$
(4) $y^2 = 1 + y\log_e\left(\frac{1+e^{-x}}{2}\right)$
Show Answer
Answer: (3)
Solution
$\frac{y^2+1}{y^2},dy = \frac{dx}{1+e^{-x}}$. Integrate: $y - \frac{1}{y} = \ln\left(\frac{e^x+1}{2}\right) + c$. Through $(0,1)$: $0 = \ln 1 + c$, $c = 0$. So $y^2 - 1 = y\log_e\left(\frac{e^x+1}{2}\right)$, i.e. $y^2 = 1 + y\log_e\left(\frac{1+e^x}{2}\right)$.
BETA
