Answer
$y=C_{1}+C_{2}e^{t}+C_{3}e^{-t}+ln(cos(t)+1)-ln(sin(t))+\frac{1}{2}e^t\int_{t_{0}}^{t}(\frac{e^{-s}}{sin(s)})ds+\frac{1}{2}e^{-t}\int_{t_{0}}^{t}(\frac{e^{s}}{sin(s)})ds$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}'''-{y}'=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}-re^{rt}=0\\\\$
$r^3-r=r(r^2-1)=0 $
$ \rightarrow\;\;\;\;\; r_{1}=0\;\;\;\;\;\;\;or\;\;\;\;,r_{2,3}=\pm 1\;\;\;\;\;\;\;\;\;\;\\\\$
$\boxed{y_{c}(t)= C_{1}+C_{2}e^{t}+C_{3}e^{-t}}$
$y_{1}=1\;\;\;\;\;,\;\;y_{2}=e^t\;\;\;\;\;\;\;,\;\;\;y_{3}=e^{-t}\;\;\;\;\;\;\;\;\;\;\;$
$W(1,e^t,e^{-t})=\begin{vmatrix}
1 & e^t & e^{-t} \\
0 & e^t & -e^{-t} \\
0 & e^t & e^{-t} \\
\end{vmatrix}\;=\;2$
$W_{1}=\begin{vmatrix}
0 & e^t & e^{-t} \\
0 & e^t & -e^{-t} \\
csc(t) & e^t & e^{-t} \\
\end{vmatrix}\;=\;-2csc(t)$
$W_{2}=\begin{vmatrix}
1 & 0 & e^{-t} \\
0 & 0 & -e^{-t} \\
0 & csc(t) & e^{-t} \\
\end{vmatrix}\;=\;e^{-t}csc(t)$
$W_{3}=\begin{vmatrix}
1 & e^t & 0 \\
0 & e^t & 0 \\
0 & e^t & csc(t) \\
\end{vmatrix}\;=\;e^tcsc(t)$
${u}'_{1}=\frac{w_{1}}{w}=-csc(t)$
$u_{1}=\int -csc(t)=ln(cot(t)+csc(t))$
${u}'_{2}=\frac{w_{2}}{w}=\frac{e^{-t}}{2sin(t)}$
$u_{2}=\int \frac{e^{-t}}{2sin(t)}= \frac{1}{2} \int_{t_{0}}^{t}(\frac{e^{-t}}{sin(t)})$
${u}'_{3}=\frac{w_{3}}{w}=\frac{e^{t}}{2sin(t)}$
$u_{3}=\int \frac{e^{t}}{2sin(t)}= \frac{1}{2} \int_{t_{0}}^{t}(\frac{e^{t}}{sin(t)}) $
$y_{p}=y_{1}u_{1}+y_{2}u_{2}+y_{3}u_{3}$
$y_{p}=ln(cos(t)+1)-ln(sin(t))+\frac{1}{2}e^t\int_{t_{0}}^{t}(\frac{e^{-s}}{sin(s)})ds+\frac{1}{2}e^{-t}\int_{t_{0}}^{t}(\frac{e^{s}}{sin(s)})ds$
$y=y_{c}+y_{p}$
$y=C_{1}+C_{2}e^{t}+C_{3}e^{-t}+ln(cos(t)+1)-ln(sin(t))+\frac{1}{2}e^t\int_{t_{0}}^{t}(\frac{e^{-s}}{sin(s)})ds+\frac{1}{2}e^{-t}\int_{t_{0}}^{t}(\frac{e^{s}}{sin(s)})ds$
