Question
If y = 5 cos x - 3 sin x, prove that $\frac{{{d^2}y}}{{d{x^2}}} + y = 0$

Answer

Let y = 5 cos x - 3 sin x ....(i)

$\therefore \frac{{dy}}{{dx}} = - 5\sin x - 3\cos x$

$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = - 5\cos x + 3\sin x$

= -(5 cos x - 3 sin x) = -y [From eq. (i)]

$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = - y$

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

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