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

Explanation

Here, $$\frac{d y}{d t}-\left(\frac{t}{1+t}\right) y=\frac{1}{1+t}$$ and $$y(0)=-1$$
which represents linear differential equation of first order.
$$
\begin{aligned}
&I F=e^{-\int\left(\frac{t}{1+t}\right) d t}=e^{-\int \frac{(t+1+1)}{1+t} d} \\
&=e^{-t-\log (1-t)} \\
&=e^{-1}(1+t)
\end{aligned}
...