Find the general solution of this differential equation.
d2ydt2+4dydt+4y=5t+3\displaystyle \frac{{{d}^{{2}}{y}}}{{{\left.{d}{t}\right.}^{{2}}}}+{4}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{y}=-{5}{t}+{3}
Because the solution is of the form
k1y1(t)+k2y2(t)+yp(t)\displaystyle {k}_{{1}}{y}_{{1}}{\left({t}\right)}+{k}_{{2}}{y}_{{2}}{\left({t}\right)}+{y}_{{p}}{\left({t}\right)} ,
please use C\displaystyle {C} and D\displaystyle {D} as the arbitrary constants. NOTE: If your first answer does not work, you should try switching the C\displaystyle {C} & D\displaystyle {D} to the alternate solutions of the homogeneous equation.