Show that \(y = \frac { 1 } { 2 } x ^ { 2 } \mathrm { e } ^ { x }\) is a solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = \mathrm { e } ^ { x }$$
Solve the differential equation \(\quad \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = \mathrm { e } ^ { x }\).
given that at \(x = 0 , y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2\).