A) \[-\frac{1}{2}\]
B) \[e+\frac{1}{2}\]
C) \[e-\frac{1}{2}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
Here,\[\frac{dy}{dt}-\left( \frac{t}{1+t} \right)y=\frac{1}{1+t}\]and\[y(0)=-1\] which represents linear differential equation of first order. \[IF={{e}^{-\int{\left( \frac{t}{1+t} \right)dt}}}={{e}^{-\int{\frac{(t+1-1)}{1+t}dt}}}\] \[={{e}^{-t+\log (1+t)}}\] \[={{e}^{-t}}(1+t)\] \[\therefore \]Required solutions is \[y(IF)=[\int{Q\cdot (IF)dt]}+C\] \[y{{e}^{-1}}(1+t)=\int{\frac{1}{1+t}{{e}^{-1}}(1+t)dt+C}\] \[=\int{{{e}^{-t}}dt+C=-{{e}^{-t}}+C}\] At \[t=0,\,\,y=-1\] \[\therefore \] \[-1(1+0)=-{{e}^{0}}+C\] \[\Rightarrow \] \[C=0\] \[\therefore \] \[y{{e}^{-t}}(1+t)=-{{e}^{-t}}\] At \[t=1\] \[y{{e}^{-1}}(1+1)=-{{e}^{-1}}\] \[\Rightarrow \] \[y=-\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
