Answer
$y=\frac{1}{2}(cosh(t)-cos(t))+\frac{1}{2}(sinh(t)-sin(t))$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
$y^{(4)}-1=0 \;\;\;\;\Rightarrow \;\;\;\; r^4e^{rt}-1=0\\\\$
$r^4-1=0 \rightarrow\;\;\;\;\; r_{1}= 1\;\;\;,\;\;\;r_{2}=-1\;\;\;or\;\;\;\;r_{3}=i \;\;\;\;\;r_{4}=-i\\\\$
So the 4 roots is: $r_{1},r_{2}=\pm 1 \;\;\;\;or\;\;\;r_{3},r_{4}=\pm i$
The general solution for complex roots is:
$y= C_{1}e^{\alpha t}cos(\beta t)+C_{2}e^{\alpha t}sin(\beta t)$
$y= Ce^{\pm t}+C_{3}cos(t)+C_{4}sin(t)$
Euler transformations:
$e^{\pm t} ={C_{1}}'cosh(t)+{C_{2}}'sinh(t)$
So;
$\boxed{y= C_{1}cosh(t)+C_{2}sinh(t)+C_{3}cos(t)+C_{4}sin(t)}$
Derivatives of the general solution;
${y}'=C_{1}sinh(t)+ C_{2}cosh(t)-C_{3}sin(t)+C_{4}cos(t)$
${y}''=C_{1}cosh(t)+C_{2}sinh(t)-C_{3}cos(t)-C_{4}sin(t)$
${y}'''=C_{1}sinh(t)+ C_{2}cosh(t)+C_{3}sin(t)-C_{4}cos(t)$
At;
$y(0)=C_{1}+C_{3}=0 $
${y}'(0)=C_{2}+ C_{4}=0 $
${y}''(0)=C_{1}- C_{3}=1 $
${y}'''(0)=C_{2}- C_{4}=1 $
$\;\;\;\;\;\Rightarrow \;\;\;\;C_{1}=\frac{1}{2}\;\;\;\;\;\;\;C_{2}=\frac{1}{2}\;\;\;\;\;\;C_{3}=\frac{-1}{2} \;\;\;\;\;\; C_{4}=\frac{-1}{2}$
$\therefore y=\frac{1}{2}(cosh(t)-cos(t))+\frac{1}{2}(sinh(t)-sin(t))$
