Higher Order Linear Differential Equations - Solving for the particular solution Given $y''' - 5y'' - y' + 5y = 3e^{-x}$, find the general solution.
I found the roots for the homogeneous solution to be 5, 1, and -1:
$$(r - 5)(r + 1)(r - 1)=0$$
$$y_h(x) = c_1e^x + c_2e^{-x} + c_3e^{5x}$$
Setting up the particular solution, I have:
$$g(x) = 3e^{-x}$$
$$y_p(x) = Axe^{-x}$$
$$y_p'(x) = Ae^{-x} - Axe^{-x} = A(1 - x)e^{-x}$$
I know I need to use the product rule to continue differentiating $y'$, but is there an easier method to do so?
 A: *

*How to differentiate $y'$ using the product rule? $y''=A(1-x)'e^{-x}+A(1-x)(e^{-x})'$, and so on.

*It is easier (IMO) to use annihilator method [1] which says:


*

*If your ODE is $L[y]=f(x)$, where $f(x)=P_k(x)e^{\alpha x}\cos(\beta x)$ or $f(x)=P_k(x)e^{\alpha x}\sin(\beta x)$ (where $P_k(x)$ is a polynomial and: $k$, $\alpha$, $\beta$ can be $0$ in order to get rid of one of these elements).

*Apply the following differential operator on both sides: $((D-\alpha)^2+\beta^2)^{k+1}$. Applying this on the right side will result in a zero.

*This will result in $y=y_h+y_p$. Since you already found $y_h$, you can identify $y_p$.

*Solve the original equation for $y_p$ in order to find the constant coefficients.


A: There is a way to get the complete solution directly.  It helps that all our roots are real.  There is a way to "factor" the left side like a polynomial.  This method attempts to get the equation into the form
$$z'+az=f(x)$$
which can be solved through the use of an integrating factor.
You have already that the characteristic polynomial is $(r-5)(r+1)(r-1)$, which I will rewrite as $(r-5)(r^2-1)$.  Now let $z=y''-y$, which you'll notice has the characteristic polynomial $r^2-1$.  Now the original equation can be rewritten.
$$y'''-y'-5y''+5y=(y''-y)'-5(y''-y)=z'-5z=3e^{-x}$$
You'll notice that the characteristic polynomial for this equation in $z$ corresponds to the other factor $r-5$.  This is of course solved my multiplying through by the integrating factor $e^{-5x}$
$$e^{-5x}z'-5e^{-5x}z=(e^{-5x}z)'=3e^{-6x}$$
$$e^{-5x}z=-\frac12e^{-6x}+k_1$$
$$z=-\frac12e^{-x}+k_1e^{-5x}$$
As long as we don't get our equation into the form
$$u'+u=f(x)$$
the $e^{-x}$ term will not cancel out prior to integration and introduce an $xe^{-x}$ term, which would make further integration more complicated.  So now we have
$$z=y''-y=-\frac12e^{-x}+k_1e^{-5x}$$
Again, we choose a substitution based on a factor of the characteristic polynomial.  To avoid the problem mentioned above, we make the substitution $u=y'+y$.  This yields
$$y''-y=(y'+y)'-(y'+y)=u'-u=-\frac12e^{-x}+k_1e^{-5x}$$
$$e^{-x}u'-e^{-x}u=(e^{-x}u)'=-\frac12e^{-2x}+k_1e^{-6x}$$
$$e^{-x}u=\frac14e^{-2x}+k_2e^{-6x}+k_3,k_2=-\frac{k_1}6$$
$$u=y'+y=\frac14e^{-x}+k_2e^{-5x}+k_3e^x$$
And finally we can solve for $y$.
$$e^xy'+e^xy=(e^xy)'=\frac14+k_2e^{-4x}+k_3e^{2x}$$
$$e^xy=\frac14x+k_4e^{-4x}+k_5e^{2x}+k_6$$
$$y=\frac14xe^{-x}+k_4e^{-5x}+k_5e^x+k_6e^{-x}$$
A: Here is an easier approach (I will give only a recipe, but it can be easily justified).
You have the characteristic polynomial, in your case this is
$$
p(r)=r^3-5r^2-r+5
$$
You need to find your particular solution to the problem with the right-hand side $3e^{-x}$ and you know that $-1$ is a root of the characteristic polynomial (multiplicity 1). A particular solution will have the form
$$
y_p(x)=\frac{3xe^{-x}}{p'(-1)}=\frac{1}{4}xe^{-x}.
$$
Added: If you need to find a particular solution to the problem with the right hand side in the form $Ae^{ax}$ and you know that $a$ is not a root of the characteristic polynomial, then a particular solution can be taken as 
$$
y_p=\frac{Ae^{ax}}{p(a)},
$$
there is no need for any derivatives.
