For any equation,
[tex]a_ny^(n)+\dots+a_1y'+a_0y=0[/tex]
assume solution of a form, [tex]e^{yt}[/tex]
Which leads to,
[tex](e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0[/tex]
Simplify to,
[tex]e^{yt}(y^3-y^2+4y-4)=0[/tex]
Then find solutions,
[tex]\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}[/tex]
For non repeated real root y, we have a form of,
[tex]y_1=c_1e^t[/tex]
Following up,
For two non repeated complex roots [tex]y_2\neq y_3[/tex] where,
[tex]y_2=a+bi[/tex]
and,
[tex]y_3=a-bi[/tex]
the general solution has a form of,
[tex]y=e^{at}(c_2\cos(bt)+c_3\sin(bt))[/tex]
Or in this case,
[tex]y=e^0(c_2\cos(2t)+c_3\sin(2t))[/tex]
Now we just refine and get,
[tex]\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}[/tex]
Hope this helps.
r3t40
