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Differential Equations

JEE Main 2021 (Online) 22th July Evening Shift

MCQ+4 / -12021
Let y = y(x) be the solution of the differential equation \(\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx\), with \(y\left( {{\pi \over 4}} \right) = 0\). Then, the value of \({(y(0) + 1)^2}\) is equal to :

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