Finding the general solution of this differential equation: $y'''' - 2y''' + 3y'' - 4y' + 2y = 0$? I am trying to find the general solution of the following differential equation: $$y'''' - 2y''' + 3y'' - 4y' + 2y = 0?$$
I don't know how to factor the characteristic equation of this. What should I do?
 A: Since the coefficients are constants try a solution of the form
$$ y(x) = e^{m}.$$ 
This leads to solving the equation $m^{4} - 2 m^{3} + 3 m^{2} - 4 m + 2 = 0$. This factors into the form $(m-1)^{2} (m^{2} + 2) = 0$ and has solutions $m \in \{ 1, 1, i \sqrt{2}, - i \sqrt{2} \}$. Now the solution is of the form
$$y(x) = A e^{x} + B x \, e^{x} + C e^{i \sqrt{2} \, x} + D e^{- i \sqrt{2} \, x}$$
or 
$$y(x) = A \, e^{x} + B \, x \, e^{x} + C \, \cos(\sqrt{2} \, x) + D \, \sin(\sqrt{2} \, x).$$
A: To factor the characteristic polynomial you use the Rational root theorem
In this case you search for hypothetical roots in the form $\frac{p}{q}$, where $p$ is an integer factor of the constant term, here $2$ and $q$ is an integer factor of the leading coefficient, which here is $1$. So for hypothetical roots you get $+2,-2,+1,-1$. You try all of them and see that $1$ is really a root. Then you use Horner's method to divide the initial polynomial by the polynomial $m-1$. In this way you get another polynomial, of degree $3$, with integer coefficients, and you again search for hypothetical rational roots. You again find that $m=1$ is a root and again apply Horner's rule to divide the degree $3$ polynomial by $m-1$ and finally get a quadratic polynomial, which you know how to factorize.
