5. Find the general solution to y'''-y''+4y'-4y = 0

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