Homogenous equation to higher order ODE Hello I have a quick question in regard to general form of the solution to

$$y^{(4)}-2y^{(3)}+y''=0$$

I had thought to find this solution we would consider

$r^{4}-2r^{3}+r^{2}=0$

which factors as $r^2(r-1)^{2}$

that is we have $r_{1,2}=0$ and $r_{3,4}=1$

So what from what I had thought I knew, I thought this implied a solution of form

$$y=c_1e^{0x}+c_2xe^{0x}+c_3e^{x}+c_4xe^{x}=c_1+c_2x+c_{3}e^{x}+c_{4}xe^{x}$$ with $c \in \mathbb{R}$

But wolfram says it is $$y(x)=e^x(c_{2}x+c_{1}-2c_{2})+c_{4}x+c_{3}$$

So where am I going wrong? Thanks all

PS: I hear it is correct, and I would by wondering about solving

$y^{(4)}-2y^{(3)}+y''=x$

Would I be able to use method of undetermined coefficients for this?

And if so, because r=0 is a double root of the equations, would by assumed form for a particular solution need to be $x^2({A_{o}x+A_{1}})$? If I did it with this, could I get a correct answer? Im  not sure exactly
you have $$y^{(4)} - 2y^{(3)} + y'' = 0 $$ if we set $u = y'',$  we get $$ u''-2u' + u = 0\to u = e^x, xe^x$$  solving $$y'' = e^x \to y' = e^x, xe^x \to y = e^x, xe^x, x^2e^x $$ and solving $$y'' = xe^x$$ you will not add anything new. but  $$y'' = 0  $$ adds $1, x$ to the basis of solutions found already. a fundamental set of solutions is $$\{1, x, e^x, xe^x \}.$$
it should be $$y \left( x \right) ={\it \_C1}\,{{\rm e}^{x}}+{\it \_C2}\,{{\rm e}^{x}
}x+{\it \_C3}+{\it \_C4}\,x
$$
for the particular solution of the second equation make the ansatz $$y_p=Ax^3+Bx^2$$