Question

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 :

• A. \(\frac{1}{2} - e\)
• B. \(e - 2\)
• C. \(2 + \frac{1}{e}\)
• D. \(\frac{1}{e} - 2\)

Hint:

Write it as a linear equation \(\frac{dy}{dx} + y\sec^2 x = \tan x\sec^2 x\) and substitute \(t = \tan x\).

Answer 1

The Correct Option is B

\(\frac{dy}{dx} + (\sec^2 x)\, y = \tan x \sec^2 x\). Integrating factor \(= e^{\int \sec^2 x\, dx} = e^{\tan x}\).
\(y e^{\tan x} = \int \tan x \sec^2 x\, e^{\tan x}\, dx\). Put \(t = \tan x\): \(\int t e^{t}\, dt = (t - 1)e^{t} + C\).
So \(y = \tan x - 1 + C e^{-\tan x}\). Using \(y(0) = 0\): \(0 = -1 + C\), so \(C = 1\).
At \(x = -\frac{\pi}{4}\), \(\tan x = -1\): \(y = -1 - 1 + e^{1} = e - 2\).
Answer: (2) \(e - 2\)