JEE PYQ: Differential Equation Question 25
Question 25 - 2020 (06 Sep Shift 1)
The general solution of the differential equation $\sqrt{1+x^2+y^2+x^2y^2} + xy\frac{dy}{dx} = 0$ is (where $C$ is a constant of integration):
(1) $\sqrt{1+y^2} + \sqrt{1+x^2} = \frac{1}{2}\log_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right) + C$
(2) $\sqrt{1+y^2} - \sqrt{1+x^2} = \frac{1}{2}\log_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right) + C$
(3) $\sqrt{1+y^2} + \sqrt{1+x^2} = \frac{1}{2}\log_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right) + C$
(4) $\sqrt{1+y^2} - \sqrt{1+x^2} = \frac{1}{2}\log_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right) + C$
Show Answer
Answer: (1)
Solution
$\sqrt{(1+x^2)(1+y^2)} + xy\frac{dy}{dx} = 0$. Separate: $\frac{y,dy}{\sqrt{1+y^2}} = -\frac{\sqrt{1+x^2}}{x},dx$. LHS: $\sqrt{1+y^2}$. RHS: using $x = \tan\theta$, we get $-\sqrt{1+x^2} + \frac{1}{2}\ln\left|\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right|$. Result: $\sqrt{1+y^2} + \sqrt{1+x^2} = \frac{1}{2}\log_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right) + C$.
BETA
