A) \[\frac{{{e}^{-2x}}}{4}\]
B) \[\frac{{{e}^{-2x}}}{4}+cx+d\]
C) \[\frac{1}{4}{{e}^{-2x}}+c{{x}^{2}}+d\]
D) \[\frac{1}{4}{{e}^{-2x}}+c+d\]
Correct Answer: B
Solution :
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{-2x}}\] On integrating, we get \[\frac{dy}{dx}=\frac{{{e}^{-2x}}}{-2}+c\] Again integrating, we get \[y=\frac{{{e}^{-2x}}}{4}+cx+d\]
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