Find the complementary solution of the following differential equation. I need to find the general solution of the following differential equation:
$$
y''' - y'' - y' + y = 4e^{-t} + 3
$$
My solution: I tried to find the complementary solution first. That is, the solution to the homogeneous equation $y''' - y'' - y' + y = 0$. If I assume that $y = e^{rt}$ then I have the following characteristic equation:
$$
r^3 - r^2 - r + 1 = 0
$$
which has the following roots
$$
(r+1)(r-1)^2 = 0 \Rightarrow \,r = -1, \,r = 1, \,r = 1
$$
Since we have repeated roots I assume that the solution is of the form $y = v(t)e^{-t}$ and I then try to figure out what the function $v(t)$ should look like. I proceeded as follows:
\begin{align}
y &= v(t)e^{-t}\\
y' &= v'(t)e^{-t} - v(t)e^{-t}\\
y'' &= v''(t)e^{-t} - 2v'(t)e^{-t} + v(t)e^{-t}\\
y''' &= v'''(t)e^{-t} - 3v''(t)e^{-t} + 3v'(t)e^{-t} - v(t)e^{-t}
\end{align}
If I substitute this back into the initial equation I get the following:
$$
v'''(t) - 4v''(t) + 4v'(t) = 0
$$
I think I need to figure out what $v(t)$ looks like based on this equation but I don't really know how. I get
$$
v(t) = v''(t) - 1/4v''(t) + c_1
$$
but this doesn't get me anywhere.
Question: How do I find the complementary solution of the initial differential equation?
 A: Idea, I would put $z= y'-y$, then we have to solve $$z''-z = 4e^{-t}+3$$
A: You’re making it more complicated than it is needed.
The complementary solution is...
$ y(t) = -c_{1}e^{-t} + c_{2}e^{t} + c_{3}te^{t} + y_{p} $
Since you do not have any IVPs, the coefficients cannot be quantified.
.
A: The complementary solution of the homogenous equation is:
$$y_c(t)=C_1e^{-t}+C_2e^t+C_3te^t.$$
The general solutions is:
$$y(t)=y_c(t)+y_p(t).$$
We will guess the particular solution as:
$$y_p(t)=Ate^{-t}+B.$$
Note: The reason for not considering $Ae^{-t}$ is it is present in the complementary solution. Therefore, we multiply it by $t$. For more examples, see here.
Now you must take the derivatives of the particular solution and equate to the right-hand side of the original equation:
$$y'_p(t)=Ae^{-t}-Ate^{-t};\\
y''_p(t)=-2Ae^{-t}+Ate^{-t};\\
y'''_p(t)=Ae^{-t}-Ate^{-t};\\
Ae^{-t}+B=4e^{-t}+3 \Rightarrow A=1; B=3.$$
Hence, the general solution is:
$$y(t)=C_1e^{-t}+C_2e^t+C_3te^t+te^{-t}+3.$$
Reference: Wikipedia answer.
