Second Order Linear

For a constant-coefficient linear equation, substituting y = e^{mx} produces the auxiliary equation, and the character of its roots dictates the solution form. Here the equation \frac{d^2y}{dx^2} + 4y = 0 gives m^2 + 4 = 0, so m^2 = -4 and m = \pm 2i, a pair of purely imaginary conjugate roots. The principle is that complex roots \alpha \pm \beta i yield solutions of the form e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x), where the real part \alpha governs exponential scaling and the imaginary part \beta governs oscillation frequency. With \alpha = 0 and \beta = 2, the exponential factor is e^0 = 1, leaving y = C_1\cos 2x + C_2\sin 2x. Option y = C_1 e^{2x} + C_2 e^{-2x} would require real roots \pm 2 from m^2 - 4 = 0. Option (C_1 + C_2 x)\cos 2x corresponds to a repeated complex root, which does not occur. Option with \cos 4x uses the wrong frequency, since \beta = 2, not 4. This is the standard JEE Advanced oscillatory-solution case. As a final check, differentiating C_1\cos 2x + C_2\sin 2x twice gives -4(C_1\cos 2x + C_2\sin 2x) = -4y, so y'' + 4y = 0 holds identically.