If y(t) is a solution of the equation $(1+t)\frac{dy}{dt}-ty=1$ and y(0) = -1 then y(1) is:

$-\frac{1}{2}$ $e+\frac{1}{2}$ $e-\frac{1}{2}$ $\frac{1}{2}$

$-\frac{1}{2}$

$\frac{dy}{dx}-\frac{t}{1+t}y=\frac{1}{1+t}$  $I.F.=e^{\frac{-t}{1+t}dt}=e^{-t+log(1+t)}=(1+t)e^{-t}$ $∴y(1+t)e^{-t}=\int\frac{1}{1+t}(1+t)e^{-t}dt⇒y(1+t)e^{-t}=-e^{-t}+c$ At t = 0, y = -1 ⇒ c = 0 $⇒ y(1+t)e^{-t}=-e^{-t}⇒y(1)=\frac{-1}{2}$