If dy/dx + 3/(cos^2 x) y = 1/(cos^2 x), x ∈ ((-π)/3, (π)/3 ) and y ((π)/4 ) = 4/3, then y (-(π)/4 ) equals

If dy/dx + 3/(cos^2 x) y = 1/(cos^2 x), x ∈ ((-π)/3, (π)/3… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 10 Jan · Shift I

\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

A.
$\frac{1}{3} + e^6$
B.
$-\frac{4}{3}$
C.
$\frac{1}{3} + e^3$
D.
$\frac{1}{3}$

Show answer and solution

Correct option: A

Solution

The ODE is dy/dx + 3sec²x · y = sec²x, a linear ODE with integrating factor e^(∫3sec²x dx) = e^(3tanx).

Multiplying: d/dx[y·e^(3tanx)] = sec²x · e^(3tanx).

Integrating: y·e^(3tanx) = (1/3)e^(3tanx) + C, so y = 1/3 + C·e^(-3tanx).

Apply y(π/4) = 4/3: 4/3 = 1/3 + C·e^(-3·1), so C = e³.

Thus y = 1/3 + e³·e^(-3tanx).

At x = -π/4: tan(-π/4) = -1, so y(-π/4) = 1/3 + e³·e^(3) = 1/3 + e⁶.

[JEE Main 2019 · 10 Jan · Shift I]

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JEE Main 2019 Mathematics question on Differential Equations