JEE Mains · Maths · STD 12 - 9. differential equations
If \(y = y ( x ), x \in\left(0, \frac{\pi}{2}\right)\) be the solution curve of the differential equation \(\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=\)\(2 e ^{-4 x }(2 \sin 2 x +\cos 2 x )\), with \(y \left(\frac{\pi}{4}\right)= e ^{-\pi}\), then \(y \left(\frac{\pi}{6}\right)\) is equal to.
- A \(\frac{2}{\sqrt{3}} e ^{-2 \pi / 3}\)
- B \(\frac{2}{\sqrt{3}} e ^{2 \pi / 3}\)
- C \(\frac{1}{\sqrt{3}} e ^{-2 \pi / 3}\)
- D \(\frac{1}{\sqrt{3}} e ^{2 \pi / 3}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{\sqrt{3}} e ^{-2 \pi / 3}\)
Step-by-step Solution
Detailed explanation
Given differential equation can be re-written as \(\frac{d y}{d x}+(8+4 \cot 2 x) y=\frac{2 e^{-4 x}}{\sin ^{2} 2 x}(2 \sin x+\cos 2 x)\) which is a linear diff. equation. \(\text { I.f. }= e ^{\int(8+4 \cot 2 x ) d x}= e ^{8 x +2 Cu (\sin 2 x )}\)…
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