JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(\frac{d y}{d x}+\frac{\sqrt{2} y}{2 \cos ^{4} x-\cos 2 x}= Xe ^{\tan ^{-1}(\sqrt{2} \cot 2 x )}, 0 < x < \pi / 2\) with \(y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{32}\). If \(y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{18} e^{-\tan ^{-1}(\alpha)}\), then the value of \(3 \alpha^{2}\) is equal to
- A \(4\)
- B \(3\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+\frac{\sqrt{2}}{2 \cos ^{4} x-\cos 2 x} y=x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}\) \(\int \frac{d x}{2 \cos ^{4} x-\cos 2 x}\) \(=\int \frac{d x}{\cos ^{4} x+\sin ^{4} x}=\int \frac{\operatorname{cosec}^{4} x d x}{1+\cot ^{4} x}\)…
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