If `y(t)` is a solution of `(1+t)(dy)/(dx)-t y=1a n dy(0)=-1` then show that `y(1)=-1/2dot`

Explanation

Given that, `(1+t)(dy)/(dt)-ty=1`
`Rightarrow (dy)/(dt)-((t)/(1+t))y=(1)/(1+t)`
Which is a linear differential equation.
On comparing it with `(dy)/(dt)+Py=Q`, we get
`P=-((t)/(1+t)),Q=(1)/(1+t)`
`IF=e^(-int(t)/(1+t)dt)=e^(-int(1-(1)/(1+t))dt=e-[t-log(1+t)])`
`e^(-t).e^(log(1+t))`
`=e^(-t)(1+t)`
The general solution is `y(t).(1+t)/(e')=int((1+1).e^(-t))/((1+t))dt+C`
`Rightarrow y(t)=(e^(-t))/(-1).(e^(t))/(1+t)+C' "where" C' =(Ce^(t))/(1+t)`
`Rightarrow y(t)=-(1)/(1+t)+C'`
When t=0 y=-1, then `-1=-1+C Rightarrow C'=0`
`y(t)=-(1)/(1+t) Rightarrow y(1)=-(1)/(2)`