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If \( y = 5 \cos x - 3 \sin x \) , prove that \( \large\frac{d^2y}{dx^2}\normalsize + y = 0\)

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Toolbox:
  • $y=f(x)$
  • $\large\frac{dy}{dx}$$=f'(x)$
  • $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
  • $\large\frac{d}{dx}$$(\sin x)=\cos x$
  • $\large\frac{d}{dx}$$(\cos x)=-\sin x$
Step 1:
$y=5\cos x-3\sin x$
Differentiating with respect to $x$
$\large\frac{dy}{dx}$$=-5\sin x-3\cos x$
Step 2:
$\large\frac{d^2y}{dx^2}$$=-5\cos x+3\sin x$
$\quad\;=-(5\cos x-3\sin x)$
$\quad\;=-y$
$\large\frac{d^2y}{dx^2}$$+y=0$
answered May 13, 2013 by sreemathi.v
 
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