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JEE Mains · Maths · STD 12 - 9. differential equations

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

  1. A \(e^{1 / 2}\)
  2. B \(e^{-1 / 2}\)
  3. C \(\mathrm{e}^{-1}\)
  4. D \(e\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{e}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\frac{d y}{d x}+2 \sin ^{2} x=1+y \cos 2 x\) \(\Rightarrow \frac{d y}{d x}+(-\cos 2 x) y=\cos 2 x\) \(\text { I.F. }=e^{\int-\cos 2 x d x}=e^{-\frac{\sin 2 x}{2}}\) Solution of \(D.E.\) \(y \cdot e^{\frac{-\sin 2 x}{2}}=\int(\cos 2 x)^{e} \frac{=\sin 2 x}{2} d x+c\)…
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