A) 0
B) \[-1\]
C) 1
D) None of these
Correct Answer: A
Solution :
Given, \[{{y}^{2}}=a{{e}^{-2x}}+\frac{2}{5}(\cos x-2\sin x)\] ...(i) On differentiating w.r.t.\[x,\]we get \[2y\frac{dy}{dx}=-2a{{e}^{-2a}}+\frac{2}{5}(-\sin x-2\cos x)\] \[\Rightarrow \] \[y\frac{dy}{dx}=-a{{e}^{-2x}}+\frac{1}{5}(-\sin x-2\cos x)\] \[\therefore \] \[y=\frac{dy}{dx}+{{y}^{2}}+\sin x\] \[=-a{{e}^{-2x}}+\frac{1}{5}(-\sin x-2\cos x)+a{{e}^{-2x}}\] \[+\frac{2}{5}(cox-2\sin x)+\sin x\] \[=-\frac{1}{5}\sin x-\frac{2}{5}\cos x+\frac{2}{5}\cos x-\frac{4}{5}\sin x\] \[+\sin x\] \[=0\]
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