JEE PYQ: Differential Equation Question 37
Question 37 - 2019 (10 Apr Shift 1)
If $y = y(x)$ is the solution of the differential equation $\frac{dy}{dx} = (\tan x - y)\sec^2 x$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that $y(0) = 0$, then $y\left(-\frac{\pi}{4}\right)$ is equal to:
(1) $e - 2$
(2) $\frac{1}{2} - e$
(3) $2 + \frac{1}{e}$
(4) $\frac{1}{e} - 2$
Show Answer
Answer: (1)
Solution
$\frac{dy}{dx} + y\sec^2 x = \tan x\sec^2 x$. I.F. $= e^{\tan x}$. Solution: $ye^{\tan x} = \int \tan x\sec^2 x\cdot e^{\tan x},dx$. Let $u = \tan x$: $= \int ue^u,du = (u-1)e^u + C = (\tan x - 1)e^{\tan x} + C$. $y(0) = 0$: $0 = -1 + C$, $C = 1$. $ye^{\tan x} = (\tan x - 1)e^{\tan x} + 1$. At $x = -\frac{\pi}{4}$: $ye^{-1} = (-2)e^{-1} + 1$, $y = -2 + e = e - 2$.
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