A) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}+2y=0\]
B) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}-2y=0\]
C) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{2}}+y=0\]
D) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-7\frac{dy}{dx}+2y=0\]
Correct Answer: A
Solution :
Given, \[y={{e}^{x}}(A\cos x+B\sin x)\] ... (i) On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}={{e}^{x}}A\cos x-{{e}^{x}}A\sin x+B{{e}^{x}}\sin x\] \[+B{{e}^{x}}\cos x\] \[\Rightarrow \]\[\frac{dy}{dx}={{e}^{x}}(A\cos x+B\sin x)-{{e}^{x}}A\sin x\] \[+B{{e}^{x}}\cos x\] \[\Rightarrow \]\[\frac{dy}{dx}=y-{{e}^{x}}A\sin x+B{{e}^{x}}\cos x\] ...(ii) Again, differentiating w.r.t.\[x,\]we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}-{{e}^{x}}A\sin x-{{e}^{x}}A\cos x-B{{e}^{x}}\sin x\] \[+B{{e}^{x}}\cos x\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}-{{e}^{x}}(A\cos x+B\sin x)\] \[-{{e}^{x}}A\sin x+B{{e}^{x}}\cos x\] \[\Rightarrow \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}-y-{{e}^{x}}A\sin x+B{{e}^{x}}\cos x\] [from Eq. (i)] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}-y+\left( \frac{dy}{dx}-y \right)\] [from Eq. (ii)] \[\Rightarrow \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2\frac{dy}{dx}-2y\] \[\Rightarrow \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}+2y=0\]
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