Can we reduce a 3rd order linear ODE to a 2nd order linear ODE? This ODE came out in my exam
$$y''' -4y'' -12y'=x^2 $$
However, I was only taught till 2nd order linear ODEs. Can we reduce this to a 2nd order or even a 1st order ODE and then use substitution to find the general solutions at the end? How does it work?
 A: We are given
$$y''' -4y'' -12y'=x^2$$
We can set
$$\begin{align} y_1' &= y' = y_2 \\ y_2' &= y'' = y_3 \\ y_3' &= y''' = 4 y'' + 12y' + x^2 = 12 y_2 + 4 y_3 + x^2 \end{align}$$
In matrix form, we have a system $Y'(x) = AY(x) + g(x)$ as
$$Y' = \begin{bmatrix} y_1' \\ y_2' \\ y_3' \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 12 & 4 \end{bmatrix}\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ x^2 \end{bmatrix}$$
Now you have a nonhomogeneous system and solve that using the standard methods or numerical methods.
Also, there are other methods to solve the original ODE, like Undetermined Coefficients, Laplace Transforms, Substitution/Integration and others.
You get the solution
$$y(x) = -\frac{1}{2} c_1 e^{-2 x}+\frac{1}{6} c_2 e^{6 x}+c_3-\frac{x^3}{36}+\frac{x^2}{36}-\frac{7 x}{216}$$
A: Part1. Yes because there's no original $y$ there. You can set $w=y'$ to get
$$w'' -4w' -12w=x^2 $$
Part2. You can't do this kind of substitution if there's an original $y$ eg
$$y''' -4y'' -12y'+7y=x^2 $$
because then you'd get something like
$$w'' -4w' -12w+7W=x^2 $$
where $W$ is some (any? Idk) antiderivative of $w$.
Part3. Anyway so go right ahead and apply what you know Gasai to
$$w'' -4w' -12w=x^2 $$
If I remember correctly, then this is

*

*Solve this $w'' -4w' -12w=0 $


*by solving this $d^2-4d-12=0$


*Solving $w'' -4w' -12w=x^2 $ given solution of $w'' -4w' -12w=0 $ or something. You know Gasai what to do.
