If $y(t)$ is a solution of $(1+t) \frac{d y}{d t}-t y=1$ and $y(0)=-1$, then $y(1)$ is equal to

$-\frac{1}{2}$

We have,

$(1+t) \frac{d y}{d t}-t y=1 \Rightarrow \frac{d y}{d t}-\frac{t}{1+t} \cdot y=\frac{1}{1+t}$       ...(i)

This is a linear differential equation with

Integrating factor = $e^{\int \frac{-t}{1+t} d t}=e^{-t+\log (1+t)}=(1+t) e^{-t}$

Multiplying (i) by integrating factor and integrating, we get

$y(1+t) e^{-t}=-e^{-t}+C \Rightarrow y(1+t)=-1+C e^t$      ....(ii)

It is given that $y(0)=-1 \Rightarrow C=0$

Putting $C=0$ in (ii), we get

$y(1+t)=-1 \Rightarrow y=-\frac{1}{1+t} \Rightarrow y(1)=-\frac{1}{2}$