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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 :
- A \(e^{1 / 2}\)
- B \(e^{-1 / 2}\)
- C \(\mathrm{e}^{-1}\)
- D \(e\)
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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