A) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2\frac{dy}{dx}-y\]
B) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2\frac{dy}{dx}-2y\]
C) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}-2y\]
D) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2\frac{dy}{dx}+y\]
Correct Answer: B
Solution :
Given \[y={{e}^{x}}A\cos x+{{e}^{x}}B\sin x\] \[\frac{dy}{dx}=A{{e}^{x}}\cos x-A{{e}^{x}}\sin x+B{{e}^{x}}\sin x+B{{e}^{x}}\cos x\] \[\frac{dy}{dx}=(A+B){{e}^{x}}\cos x+(B-A){{e}^{x}}\sin x\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=(A+B){{e}^{x}}\cos x-{{e}^{x}}\sin x(A+B)\] \[+(B-A){{e}^{x}}\sin x+(B-A){{e}^{x}}\cos x\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2B{{e}^{x}}\cos x-2A{{e}^{x}}\sin x\]. Hence \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2\frac{dy}{dx}-2y\].
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