JEE PYQ: Differential Equation Question 43
Question 43 - 2019 (10 Jan Shift 1)
If $\frac{dy}{dx} + \frac{3}{\cos^2 x}y = \frac{1}{\cos^2 x}$, $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$, and $y\left(\frac{\pi}{4}\right) = \frac{4}{3}$, then $y\left(-\frac{\pi}{4}\right)$ equals:
(1) $\frac{1}{3} + e^6$
(2) $\frac{1}{3}$
(3) $-\frac{4}{3}$
(4) $\frac{1}{3} + e^3$
Show Answer
Answer: (1)
Solution
I.F. $= e^{3\tan x}$. Solution: $ye^{3\tan x} = \int \sec^2 x\cdot e^{3\tan x},dx = \frac{1}{3}e^{3\tan x} + C$. Using $-\frac{1}{3}\ln|1-3y| = \tan x + C$ approach: at $x = \frac{\pi}{4}$, $y = \frac{4}{3}$: $-\frac{1}{3}\ln 3 = 1 + C$. At $x = -\frac{\pi}{4}$: $-\frac{1}{3}\ln|1-3y| = -1-\frac{1}{3}\ln 3 + 1$… From the solution images: $y = \frac{1}{3} + Ce^{-3\tan x}$. $\frac{4}{3} = \frac{1}{3} + Ce^{-3}$, $C = e^3$. At $x = -\frac{\pi}{4}$: $y = \frac{1}{3} + e^3 \cdot e^3 = \frac{1}{3} + e^6$.
BETA
