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JEE Main 2022 Mathematics question

MathematicsDifferential EquationJEE Main 2022

Let y = y(x) be the solution of the differential equation dy/dx + (Extra open brace or missing close brace\sqrt{2y})/(2\cos ^{4}x - \cos 2x) = x e^{\tan ^{-1}(\sqrt{2} cot 2x)}, with . If , then the value of is equal to .

Answer: 2

Solution

Since , the equation is dy/dx + (Extra open brace or missing close brace2\sqrt{2}/(1 + \cos ^{2}2x))y = x e^{\tan ^{-1}(\sqrt{2} cot2x)}. The integrating factor is Extra open brace or missing close bracee^{\int 2\sqrt{2} dx/(1 + Extra close brace or missing open brace\cos ^{2}2x)} = e^{\tan ^{-1}(\tan 2x/\sqrt{2})}. Since the product of cot2x and is 1,

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Question ID VK-035337