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JEE Main 2019 · 10 Apr · Shift I

JEE Main 2019 Mathematics question on Differential Equations

If y=y(x)y = y(x) is the solution of the differential equation dydx=(tan⁡x−y)sec⁡2x\frac{dy}{dx} = (\tan x - y)\sec^2 x, x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), such that y(0)=0y(0) = 0, then y(−π4)y\left(-\frac{\pi}{4}\right) is equal to
  1. A.
    14−2\frac{1}{4} - 2
  2. B.
    12−e\frac{1}{2} - e
  3. C.
    2+1e2 + \frac{1}{e}
  4. D.
    e−2e - 2
Show answer and solution

Correct option: D

Solution

The equation is linear: dy/dx + y·sec²x = tan(x)·sec²x.

Integrating factor = e^(∫sec²x dx) = e^(tan x).

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

Integrate the right side using substitution t = tan x:
∫t·et dt = et(t - 1) + C = e^(tan x)(tan x - 1) + C.

So y·e^(tan x) = e^(tan x)(tan x - 1) + C.

Apply y(0) = 0: 0·e⁰ = e⁰(0 - 1) + C → C = 1.

Thus y = (tan x - 1) + e^(-tan x).

At x = -π/4: tan(-π/4) = -1, so y = (-1 - 1) + e^(1) = -2 + e = e - 2.

[JEE Main 2019 · 10 Apr · Shift I]

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