Can't find particular integral of a differential equation I have been trying to solve this differential equation using Method of undetermined coefficients and all my guesses are incorrect and not working :
$$y''' + 3y'' + 3y' + y = 30e^{-x}$$
When I try to find the Particular Integral, the whole LHS is becoming 0. So I can't get any value for 'A'. Can anyone solve this differential equation for me? Or just the particular integral is enough
I stared with the guess that it could be $Ae^{-x}$. But it didn't work.  When i looked online, i came across an article, it said that i have to add an extra term x^s, where s is the smallest positive integer that renders all summands of a solution independent of the homogeneous solutions. Now i didn't exactly understand what they meant. But like in few examples they gave, i tried with just t first, then i tried $t^2$. That also didn't work. Then I tried $t^3$, and then got $\frac{5}{x-1}$ as a value for A. I don't think that's the right answer. Is there any shortcut to guessing the correct one? I know that I can try doing it with the method of variation of parameters but in our exams sometimes they specifically say that you have to use this specific method..Can anyone help?
 A: The key to solving this equation is understanding what exactly the differential operator on the LHS of the equation is doing. Let us rewrite $y''' + 3y'' + 3y' + y = 30e^{-x}$ as $(D^3)[y] + (3D^2)[y] + (3D)[y] + I[y] = 30e^{-x}$, where $D$ is the derivative operator, and $I$ is the identity operator. Think of these operators as being functions of real-variable functions. Then $(D^3)[y] + (3D^2)[y] + (3D)[y] + I[y] = (D^3 + 3D^2 + 3D + I)[y] = 30e^{-x}$. The next part is key, but this is something that, at first glance, looks like an invalid manipulation of symbols. Essentially, you can say $D^3 + 3D^2 + 3D + I = (D + I)^3 = (D + I)(D + I)(D + I)$. This does not look correct, but it is correct, and the reason it works is because $DI = ID$. In general, operators do not commute, but in this particular case, these two do commute, and this is the reason this works. As such, you have $(D + I)(D + I)(D + I)[y] = 30e^{-x}$. To solve this, it is now straightforward: just do a bunch of substitutions. Let $(D + I)[y] := y0$ and let $(D + I)[y0] = y1$, and we have that $(D + I)[y1] = 30e^{-x}$. Each of these is a first-order linear differential equation. For the latter, we have $(y1)' + y1 = 30e^{-x}$, which is equivalent to $e^x \cdot (y1)' + e^x \cdot y1 = 30$, which is equivalent to $[e^x \cdot y1]' = 30$. Hence $e^x \cdot y1 = 30x + A$, or $y1 = 30xe^{-x} + Ae^{-x}$. Thus $(D + I)[y0] = 30xe^{-x} + Ae^{-x} = (30x + A)e^{-x}$, equivalent to $(y0)' + y0 = (30x + A)e^{-x}$. You know the drill. This will solve to $e^x \cdot y0 = 15x^2 + Ax + B$. So $(D + I)[y] = (15x^2 + Ax + B)e^{-x}$, equivalent to $y' + y = (15x^2 + Ax + B)e^{-x}$. For the last time, you use the integrating factor to get $e^x \cdot y = 5x^3 + A/2 \cdot x^2 + Bx + C$, which is equuivalent to $y = (5x^3 + \frac{A}{2} x^2 + Bx + C)e^{-x}$.
The key idea was to use the factorization of the linear operator to turn the equation into multiple first order equations. By the way, this method can be proven rigorously, and will always give the correct answer whenever the coefficients of the equations are constants. So this works even without repeated factors. This method is also the explanation for why those polynomial factors appear only in equations where the characteristic polynomial has a repeated root.
A: Choose as a particular solution $y_p(x)=\alpha x^{\beta}e^{-x}$ and after substituting into the complete ODE we have
$$
\alpha  (\beta -2) (\beta -1) \beta  x^{\beta }-30 x^3 = 0\Rightarrow \cases{\alpha = 5\\ \beta = 3}
$$
so we have $y_p(x) = 5x^3 e^{-x}$
A: $$y''' + 3y'' + 3y' + y = 30e^{-x}$$
$$(y''' + 3y'' + 3y' + y )e^{x}=30$$
Is simply equivalent to:
$$(ye^{x})'''=30$$
$$z'''=30$$
This is easy to integrate.
$$z'''_p=A \implies z''_p=Ax \implies z'_p=Bx^2 \implies z_p=Cx^3$$
This should be your guess.
