Question

Let \(y = y(x)\) be the solution of the differential equation \(\operatorname{cosec}^2 x\, dy + 2\, dx = (1 + y\cos 2x)\operatorname{cosec}^2 x\, dx\), with \(y\left(\frac{\pi}{4}\right) = 0\). Then, the value of \((y(0)+1)^2\) is equal to :

• A. \(e^{1/2}\)
• B. \(e^{-1/2}\)
• C. \(e^{-1}\)
• D. \(e\)

Hint:

Divide by \(\operatorname{cosec}^2 x\) to get a linear equation and substitute \(u = y+1\).

Answer 1

The Correct Option is C

Divide by \(\operatorname{cosec}^2 x\): \(dy + 2\sin^2 x\, dx = (1 + y\cos 2x)dx\).
So \(\frac{dy}{dx} = 1 - 2\sin^2 x + y\cos 2x = \cos 2x + y\cos 2x = (1+y)\cos 2x\).
Separate variables: \(\frac{dy}{1+y} = \cos 2x\, dx \Rightarrow \ln|1+y| = \frac{\sin 2x}{2} + k\).
At \(x = \frac{\pi}{4}\), \(y = 0\): \(0 = \frac{1}{2} + k \Rightarrow k = -\frac{1}{2}\).
At \(x = 0\): \(\ln|1+y(0)| = 0 - \frac{1}{2}\), so \(1 + y(0) = e^{-1/2}\).
\((y(0)+1)^2 = e^{-1}\).
Answer: (3) \(e^{-1}\)