If y=5 -3 , prove that d^2y dx^2 +y=0 - Vidyadip

Question

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

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Answer

Let $\text{y}=5\cos\text{x}-3\cos\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-5\sin\text{x}-3\cos\text{x}$
$\Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-5\cos\text{x}+3\sin\text{x}=-(5\cos\text{x}-3\sin\text{x})=-\text{y}$
$\Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\ \Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$

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