Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{First order}} \cr
& \left( {\text{b}} \right){\text{Second order}} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)2\frac{{dy}}{{dx}} + y = x - 1;{\text{ }}y = c{e^{ - x/2}} + x - 3 \cr
& {\text{The differential equation }}2\frac{{dy}}{{dx}} + y = x - 1{\text{ is the first order}} \cr
& {\text{Checking}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {c{e^{ - x/2}} + x - 3} \right] \cr
& \frac{{dy}}{{dx}} = - \frac{c}{2}{e^{ - x/2}} + 1,{\text{ then}} \cr
& 2\left( { - \frac{c}{2}{e^{ - x/2}} + 1} \right) + c{e^{ - x/2}} + x - 3 = x - 1 \cr
& - c{e^{ - x/2}} + 2 + c{e^{ - x/2}} + x - 3 = x - 1 \cr
& x - 1 = x - 1 \cr
& y = c{e^{ - x/2}} + x - 3{\text{ is a solution}}{\text{.}} \cr
& \cr
& \left( {\text{b}} \right)y'' - y = 0;{\text{ }}y = {c_1}{e^t} + {c_2}{e^{ - t}} \cr
& {\text{The differential equation }}y'' - y = 0{\text{ is the second order}} \cr
& {\text{Checking}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{c_1}{e^t} + {c_2}{e^{ - t}}} \right] \cr
& \frac{{dy}}{{dx}} = {c_1}{e^t} - {c_2}{e^{ - t}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = {c_1}{e^t} + {c_2}{e^{ - t}},{\text{ then}} \cr
& {c_1}{e^t} + {c_2}{e^{ - t}} - {c_1}{e^t} - {c_2}{e^{ - t}} = 0 \cr
& 0 = 0 \cr
& y = {c_1}{e^t} + {c_2}{e^{ - t}}{\text{ is a solution}}{\text{.}} \cr} $$