If $$y(t)$$ is a solution of $$\left( {1 + t} \right){{dy} \over {dt}} - ty = 1$$ and $$y\left( 0 \right) = - 1,$$ then $$y(1)$$ is equal to

If $${x^2} + {y^2} = 1,$$ then

A solution of the differential equation
$${\left( {{{dy} \over {dx}}} \right)^2} - x{{dy} \over {dx}} + y = 0$$ is

The order of the differential equation whose general solution is given by
$$y = \left( {{C_1} + {C_2}} \right)\cos \left( {x + {C_3}} \right) - {C_4}{e^{x + {C_5}}},$$ where
$${C_1},{C_2},{C_3},{C_4},{C_5},$$ are arbitrary constants, is