A) \[4-\frac{2}{y}-\frac{{{e}^{\frac{1}{y}}}}{e}\]
B) \[3-\frac{1}{y}+\frac{{{e}^{\frac{1}{y}}}}{e}\]
C) \[1+\frac{1}{y}-\frac{{{e}^{\frac{1}{y}}}}{e}\]
D) \[1-\frac{1}{y}+\frac{{{e}^{\frac{1}{y}}}}{e}\]
Correct Answer: C
Solution :
\[\frac{dx}{dy}+\frac{x}{{{y}^{2}}}=\frac{1}{{{y}^{3}}}\] I.F.\[={{e}^{\int_{{}}^{{}}{\frac{1}{{{y}^{2}}}dy}}}={{e}^{-\frac{1}{y}}}\] so\[x.{{e}^{-\frac{1}{y}}}=\int_{{}}^{{}}{\frac{1}{{{y}^{3}}}{{e}^{-\frac{1}{y}}}dy}\] Let\[\frac{-1}{y}=t\] \[\Rightarrow \]\[\frac{1}{{{y}^{2}}}dy=dt\] \[\Rightarrow \]\[I=-\int_{{}}^{{}}{t{{e}^{t}}dt}={{e}^{t}}-t{{e}^{t}}\] \[={{e}^{-\frac{1}{y}}}+\frac{1}{y}{{e}^{-\frac{1}{y}}}+c\] \[\Rightarrow \]\[=x{{e}^{-\frac{1}{y}}}={{e}^{-\frac{1}{y}}}+\frac{1}{y}{{e}^{-\frac{1}{y}}}+c\] \[\Rightarrow \]\[x=1+\frac{1}{y}+c.{{e}^{1/y}}\] since\[y(1)=1\] \[\therefore \]\[c=-\frac{1}{e}\] \[\Rightarrow \]\[x=1+\frac{1}{y}-\frac{1}{e}.{{e}^{1/y}}\]
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