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 Q11    If   y = 5 \cos x - 3 \sin x  prove that \frac{d^2y}{dx^2}+y = 0
 

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Given function is
y = 5 \cos x - 3 \sin x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}=-5\sin x-3\cos x
Now, the second-order derivative is 
\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}=-5\cos x+3\sin x
Now, 
\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0
Hence proved 
 

Posted by

Gautam harsolia

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