A) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+y=0\]
B) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-y=0\]
C) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+(a+b)y=0\]
D) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=(a+b)y\]
E) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=(a-b)y\]
Correct Answer: A
Solution :
\[\because \] \[y=a\cos x+b\sin x\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=-a\sin x+b\cos x\] Again differentiating, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-a\cos x-b\sin x\] \[=-(a\cos x+b\sin x)=-y\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+y=0\]
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