If $y(t)$ is a solution of $(1 + t)\frac{{dy}}{{dt}} - ty = 1$ and
$y(0) = - 1,$ then show that $y(1) = - \frac{1}{2}$.

Given that,

$(1 + t)\frac{{dy}}{{dt}} - ty = 1$
$\Rightarrow$ $\frac{{dy}}{{dt}} - \left( {\frac{t}{{1 + t}}} \right)y = \frac{1}{{1 + t}}$

which is a linear differential equation.
On comparing it with

$\frac{{dy}}{{dt}} + Py = Q,$

we get
$P = - \left( {\frac{t}{{1 + t}}} \right),Q = \frac{1}{{1 + t}}$

${\rm{IF}} = {e^{ - \int {\frac{t}{{1 + t}}} dt}} = {e^{ - \int {\left( {1 - \frac{1}{{1 + t}}} \right)} dt = {e^{ - [t - \log (1 + t)]}}}}$

$= {e^{ - t}} \cdot {e^{\log (1 + t)}}$
$= {e^{ - t}}(1 + t)$

The general solution is
$y(t) \cdot \frac{{(1 + t)}}{{{e^t}}} = \int {\frac{{(1 + t) \cdot {e^{ - t}}}}{{(1 + t)}}} dt + C$

$\Rightarrow$ $y(t) = \frac{{{e^{ - t}}}}{{( - 1)}} \cdot \frac{{{e^t}}}{{1 + t}} + {C^\prime },$

where ${C^\prime } = \frac{{C{e^t}}}{{1 + t}}$
$\Rightarrow$ $y(t) = - \frac{1}{{1 + t}} + {C^\prime }$

When $t = 0$ and $y = - 1$,

then
$- 1 = - 1 + {C^\prime } \Rightarrow {C^\prime } = 0$

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