Show that the function of $y = A sin 2 x + B cos 2 x$ satisfies the differential equation
$\frac{d^{2} y}{d x^{2}} + 4 y = 0$

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Solution

Given function,

$y = A sin 2 x + B cos 2 x$ …….. (1)

Differentiating this equation with respect to $x$ and we get,

$\frac{d y}{d x} = A \frac{d}{d x} sin 2 x + B \frac{d}{d x} cos 2 x$

$\frac{d y}{d x} = A 2 cos 2 x + B (- 2 sin 2 x)$

$\frac{d y}{d x} = 2 A cos 2 x - 2 B sin 2 x$

Again Differentiating this equation with respect to $x$ and we get,

$\frac{d^{2} y}{d x^{2}} = 2 A \frac{d}{d x} cos 2 x - 2 B \frac{d}{d x} sin 2 x$

$\frac{d^{2} y}{d x^{2}} = 2 A (- 2 sin 2 x) - 2 B (2 cos 2 x)$

$\frac{d^{2} y}{d x^{2}} = - 4 A sin 2 x - 4 B cos 2 x$

$\frac{d^{2} y}{d x^{2}} = - 4 (A sin 2 x - B cos 2 x)$

$\frac{d^{2} y}{d x^{2}} = - 4 y$ by equation (1)

$\frac{d^{2} y}{d x^{2}} + 4 y = 0$

Hence proved.

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