Question:
If $y=5 \cos x-3 \sin x$, prove that $\frac{d^{2} y}{d x^{2}}+y=0$
Solution:
It is given that, $y=5 \cos x-3 \sin x$
Then
$\frac{d y}{d x}=\frac{d}{d x}(5 \cos x)-\frac{d}{d x}(3 \sin x)=5 \frac{d}{d x}(\cos x)-3 \frac{d}{d x}(\sin x)$
$=5(-\sin x)-3 \cos x=-(5 \sin x+3 \cos x)$
$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[-(5 \sin x+3 \cos x)]$
$=-\left[5 \cdot \frac{d}{d x}(\sin x)+3 \cdot \frac{d}{d x}(\cos x)\right]$
$=-[5 \cos x+3(-\sin x)]$
$=-[5 \cos x-3 \sin x]$
$=-y$
$\therefore \frac{d^{2} y}{d x^{2}}+y=0$
Hence, proved.
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