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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The differential equation for which $y=a\cos x+b\sin x$ is a solution is \[(A)\;\frac{d^2y}{dx^2}+y=0 \quad (B)\;\frac{d^2y}{dx^2}-y=0 \quad (C)\;\frac{d^2y}{dx^2}+(a+b)y=0 \quad (D)\;\frac{d^2y}{dx^2}+(a+b)y=0\]

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Toolbox:
  • A differential equation will be of $n^{th}$ order if its equation has $'n'$ arbitrary constants
  • $\large\frac{d}{dx}$$(\sin x)=\cos x$ and $\large\frac{d}{dx}$$(\cos x)=-\sin x$
Given $ y=a \cos x+b \sin x$
Differentiate w.r.t x on both sides,
$\large\frac{dy}{dx}$$=-a\sin x+b \cos x$
Again differentiate w.r.t x on both sides,
$\large\frac{d^2y}{dx^2}$$=-a\cos x-b \sin x$
$=-[a \cos x+b\sin x]$
But $ a \cos x+b \sin x=y$
Hence $\large\frac{d^2y}{dx^2}=-y$
$\large\frac{d^2y}{dx^2}+y=0$
Hence the correct option is $A$
answered May 20, 2013 by meena.p
 

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