Question #4886a

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Answer 1 · 2018-01-24T22:25:05

See below.

$y = p cos (2 x) + q sin (2 x)$ $y = pcos(2x)+qsin(2x)$

First differentiate:

$\frac{d y}{d x} = - 2 p sin (2 x) + 2 q cos (2 x)$ $dy/dx=-2p sin(2x)+2qcos(2x)$

$\frac{d^{2} y}{{d x}^{2}} = - 4 p cos (2 x) - 4 q sin (2 x)$ $(d^2y)/(dx^2)=-4p cos(2x)-4qsin(2x)$

Now substitute into:

$\frac{d^{2} y}{{d x}^{2}} + 4 y$ $(d^2y)/(dx^2)+4y$

$= - 4 p sin (2 x) - 4 q cos (2 x) + 4 (p sin (2 x) + q cos (2 x))$ $=-4p sin(2x)-4qcos(2x)+4(p sin(2x)+qcos(2x))$

$= - 4 p sin (2 x) - 4 q cos (2 x) + 4 p sin (2 x) + 4 q cos (2 x)$ $=-4p sin(2x)-4qcos(2x)+4p sin(2x)+4qcos(2x)$

$= - 4 p sin (2 x) + 4 p sin (2 x) - 4 q cos (2 x) + 4 q cos (2 x)$ $=-4p sin(2x)+4p sin(2x)-4qcos(2x)+4qcos(2x)$

$= 0$ $=0$