Answer
$y=2-2cos(t)+sin(t)
$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}'''+{y}'=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}+re^{rt}=0\\\\$
$r(r^2+1)=0 \rightarrow\;\;\;\;\; r_{1}= 0\;\;\;\;\;or\;\;\;r_{2}=i\;,\;\;r_{3}=-i \;\;\;\;\;\;\\\\$
So the 3 roots are:$\;\;\; r_{1}=0 \;\;\;,\;\;r_{2},r_{3}=\pm i $
The general solution for complex roots is:
$y= C_{1}e^{\alpha t}cos(\beta t)+C_{2}e^{\alpha t}sin(\beta t)$
$y= C_{1}+C_{2}cos(t)+C_{3}sin(t)$
Derivatives of the general solution;
${y}'=-C_{2}sin(t)+C_{3}cos(t)$
${y}''=-C_{2}cos(t)-C_{3}sin(t)$
At;
$y(0)=C_{1}+C_{2}=0 \;\;\;\;\;\Rightarrow \;\;\;\;\;\;\;C_{1}=-C_{2}$
${y}'(0)=C_{3}=1$
${y}''(0)=-C_{2}=2 \;\;\;\;\;\;\rightarrow \;\;\;\;\;C_{1}=2$
$\therefore \;\;\;\;\;\;\;\;\;y=2-2cos(t)+sin(t)
$
