Higher Order Method of Undetermined Coefficients i'm working on a textbook problem trying to find the general solution of the given differential equation.
The question is $$y''''-y=3t+cost$$ I started by finding the characteristic equation to which i get $$r^4-1+0$$
From this i have my roots are $r=1, r=1, r=-1, r=-1$ so i thought that my $$y_c(t) = c_1e^t+c_2te^t+c_3e^{-t}+c_4te^{-t}$$
Why is this wrong?
 A: The solution to your differential comes from solving the homogenous portion of your equation, and then the nonhomogeneous. \begin{equation}y^{(4)}-y=3t+\cos(t) \end{equation}
Now to state the homogenous, and auxiliary equation: \begin{align}y^{(4)}-y&=0 \\ m^4-1&=0 \end{align} Which factors nicely to the following equation: \begin{align}(m^2+1)(m^2-1)&=0\\ (m^2+1)(m+1)(m-1)&=0 \end{align} Setting each term equal to zero one gets the following: \begin{align}m^2&=1\\ m&=-1 \\ m&=1\end{align} The first equation produces imaginary solutions so one can write the roots as the following: \begin{equation} m=\pm i, \pm 1\end{equation}. Roots one can begin to designate the homogenous solutions to the equation: \begin{align}y_c&=c_1 e^t+c_2 e^{-t}+e^{0t}(c_3\cos(t)+c_4\sin(t))\\ y_c&=c_1 e^t+c_2 e^{-t}+c_3 \cos(t) + c_4 \sin(t) \end{align}
Now one can begin to tackle the non-homogenous part using Undetermined coefficients. Therefore for our guesswork we are are going to guess the following functions for $y_p$. \begin{equation}y_p=At+B+C\sin(t)+D\cos(t)\end{equation} Then we will backtrack and substitute $y$ for $y_p$. \begin{equation}y^{(4)}_p-y_p=3t+\cos(t) \end{equation}
Now we take the four derivatives and achieve the following equation: \begin{equation} c\sin(t)+d\cos(t)-at-b-c\sin(t)-d\cos(t)=3t+\cos(t)\end{equation} From this equation one can notice that the $c$, and $d$ terms cancel out, therefore one has to increase in linear independence the sines and cosines. therefore, the particular solution looks as the following now: \begin{equation}y_p=At+B+Ct\sin(t)+Dt\cos(t) \end{equation} Now our set up looks as the following formulation: \begin{align}(ct+4d)\sin(t)+(dt-4t)\cos(t)-at-b-ct\sin(t)-dt\cos(t)&=3t-\cos(t) \\ 4d\sin(t)-4ct\cos(t)-at-a&=3t-\cos(t)\\-a&=3\\-4c&=1\\b&=0\\d&=0\end{align} Then we get the following: \begin{equation}y_p=-3t-\frac{1}4t\sin(t)  \end{equation} Therefore the whole solution to your problem is the following:\begin{equation} y=c_1 e^t+c_2 e^{-t}+c_3 \cos(t) + c_4 \sin(t)-3t-\frac{1}4t\sin(t)\end{equation}
