A) zero
B) \[\frac{-16Gm}{r}\]
C) \[\frac{-16Gm}{3r}\]
D) \[\frac{-3Gm}{16r}\]
Correct Answer: C
Solution :
\[-{{e}^{-2x}}\,=t\Rightarrow \,\frac{dt}{dy}+\frac{2t}{y}=\frac{2}{{{y}^{3}}}\] \[\therefore \,\,t.{{y}^{2}}\,=\int_{{}}^{{}}{\frac{2}{{{y}^{3}}}.\,{{y}^{2}}dy+C}\] \[\Rightarrow -{{e}^{-2x}}\,.{{y}^{2}}\,=2\ell n\,y\,+C\] \[y(0)=1\] \[\Rightarrow \,C=-1\]
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