A) \[\frac{-1}{2}\]
B) \[e+\frac{1}{2}\]
C) \[e-\frac{1}{2}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
\[(1+t)\frac{dy}{dt}-ty=1\] \[\frac{dy}{dt}-\left( \frac{t}{1+t} \right)y=\frac{1}{1+t}\] \[I.F.={{e}^{-\int{\frac{t}{1+t}dt}}}={{e}^{-t+\ln \,(t+1)}}(t+1){{e}^{-t}}\] \[\therefore \,\,y(t+1){{e}^{-t}}=\int{\frac{1}{(t+1)}}\cdot (t+1){{e}^{-1}}dt+C\] \[y(t+1){{e}^{-t}}=-{{e}^{-t}}+C\] At \[t=0,\,y=-1\] \[\therefore C=0\] \[\therefore \,y=\frac{-1}{t+1}\Rightarrow y(1)=\frac{-1}{2}\]
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