Answer
$\;\;\;\;\;\;C_{1}=2\;\;\;\;\;\;\;C_{2}=\frac{7}{8}\;\;\;\;\;\;\;\;\;C_{3}=\frac{-7}{8}\;\;\;\;\;\;\;\;\;C_{4}=\frac{1}{2}$
$y(t)=2cos(t)+\frac{7}{8}sin(t)-\frac{7}{8}tcos(t)+\frac{1}{2}tsin(t)-\frac{1}{8}t^2sin(t)$
Work Step by Step
The solution of the differential equation was found out in problem (6) :
$y(t)=C_{1}cos(t)+C_{2}sin(t)+C_{3}tcos(t)+C_{4}tsin(t)-\frac{1}{8}t^2sin(t)$
${y}'=-C_{1}sin(t)+C_{2}cos(t)+C_{3}cos(t)-C_{3}tsin(t)+C_{4}sin(t)+C_{4}tcos(t)-\frac{1}{4}tsin(t)-\frac{1}{8}t^2cos(t)$
${y}''=-C_{1}cos(t)-C_{2}sin(t)-2C_{3}sin(t)-C_{3}tcos(t)+2C_{4}cos(t)-C_{4}tsin(t)-\frac{1}{2}tcos(t)-\frac{1}{4}sin(t)+\frac{1}{8}t^2sin(t)$
${y}'''=C_{1}sin(t)-C_{2}cos(t)-3C_{3}cos(t)+C_{3}tsin(t)-3C_{4}sin(t)-C_{4}tcos(t)-\frac{3}{4}cos(t)+\frac{3}{4}tsin(t)+\frac{1}{8}t^2cos(t)$
$y(0)=C_{1}=2$
${y}'(0)=C_{2}+C_{3}=0$
${y}''(0)=-C_{1}+2C_{4}=-1$
${y}'''(0)=-C_{2}-3C_{3}-\frac{3}{4}=1$
So; $\;\;\;\;\;\;C_{1}=2\;\;\;\;\;\;\;C_{2}=\frac{7}{8}\;\;\;\;\;\;\;\;\;C_{3}=\frac{-7}{8}\;\;\;\;\;\;\;\;\;C_{4}=\frac{1}{2}$
$y(t)=2cos(t)+\frac{7}{8}sin(t)-\frac{7}{8}tcos(t)+\frac{1}{2}tsin(t)-\frac{1}{8}t^2sin(t)$
