JEE PYQ: Differential Equation Question 38
Question 38 - 2019 (10 Apr Shift 2)
Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y\tan x = 2x + x^2\tan x$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $y(0) = 1$. Then:
(1) $y\left(\frac{\pi}{4}\right) + y\left(-\frac{\pi}{4}\right) = \frac{\pi^2}{2} + 2$
(2) $y\left(\frac{\pi}{4}\right) + y\left(-\frac{\pi}{4}\right) = -\sqrt{2}$
(3) $y\left(\frac{\pi}{4}\right) - y\left(-\frac{\pi}{4}\right) = \sqrt{2}$
(4) $y’\left(\frac{\pi}{4}\right) - y’\left(-\frac{\pi}{4}\right) = \pi - \sqrt{2}$
Show Answer
Answer: (4)
Solution
I.F. $= \sec x$. Solution: $y\sec x = \int (2x + x^2\tan x)\sec x,dx = x^2\sec x + c$. $y(0) = 1$: $c = 1$. So $y = x^2 + \cos x$. $y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{16} + \frac{1}{\sqrt{2}}$, $y\left(-\frac{\pi}{4}\right) = \frac{\pi^2}{16} + \frac{1}{\sqrt{2}}$. $y’ = 2x - \sin x$. $y’\left(\frac{\pi}{4}\right) = \frac{\pi}{2} - \frac{1}{\sqrt{2}}$, $y’\left(-\frac{\pi}{4}\right) = -\frac{\pi}{2} + \frac{1}{\sqrt{2}}$. Difference $= \pi - \sqrt{2}$.
BETA
