The differential equation should be
x ¨ + 1 t 2 + 2 t + 2 x ˙ + 1 t 2 + 4 t + 8 x = 0 {\displaystyle {\ddot {x}}+{\frac {1}{t^{2}+2t+2}}{\dot {x}}+{\frac {1}{t^{2}+4t+8}}x=0}
Return to errata.