A) \[y=\frac{1}{4}{{e}^{-2x}}+\frac{cx}{2}+d\]
B) \[y=\frac{1}{4}{{e}^{-2x}}+cx+d\]
C) \[y=\frac{1}{4}{{e}^{-2x}}+c{{x}^{2}}+d\]
D) \[y=\frac{1}{4}{{e}^{-2x}}+c{{x}^{3}}+d\]
Correct Answer: B
Solution :
Given equation\[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{-2x}}\] On integrating both sides \[\int{\frac{{{d}^{2}}y}{d{{x}^{2}}}dx}=\int{{{e}^{-2x}}}dx\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{{{e}^{-2x}}}{-2}+c\] Again integrating, we get \[y=\frac{{{e}^{-2x}}}{4}+cx+d\]
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