Find a general solution and check if it is independent for $y''=0$?

Question

I usually know how to solve these type of ODEs and that the way to find the general solution is that $y=C_1e^{xr1}+C_2e^{xr2}$. But in this case when we do $r^2=0$ then $r$ can only be $0$ and in the notes $y_1=e^{0x}$ which is $1$ but the $y_2=xe^{0x}$ and I don't know where this $x$ comes from. In some cases this "$x$" just appears out of nowhere. I'm probably missing some theory here.

Answer 1

integrate twice ... $$y''=0$$ Integrate $$y'=K_1$$ Integrate again $$y=\int K_1dx=K_1\int dx$$ $$y=K_1x+K_2$$ Note that $e^{0x}=1$ $$y=K_1xe^{0x}+K_2e^{0x}$$

or you can use the reduction of order once you know a particular solution $y_2=y_1v(x)=Kv(x)$ Plug that in your equation to get $y_2$

Answer 2

Let's start with a general approach:

Consider $(D-r)^2y=0$, where $D\equiv\frac{d}{dx}$. Then the general solution is $y(x)=c_1e^{rx}+c_2e^{rx}=(c_1+c_2)e^{rx}=ce^{rx}$, where $c=c_1+c_2$. But this is not a general solution, since the number of arbitrary constant is now $1$ and not $2$.

Now $(D-r)^2y=0$ implies $(D-r)v=0$ where $v=(D-r)y$.

Thus $v=c_1e^{rx}$ i.e., $$(D-r)y=c_1e^{rx}.$$ By solving $$y=c_1xe^{rx}+c_2e^{rx}.$$