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
 A: 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 $$\{1, x, e^x, xe^x \}.$$
A: For the equation $y^{(4)} - a y^{(3)} + y^{(2)} = c_{0}x$ the solution is as follows. Integrate twice to obtain $y^{(2)} - a y^{(1)} + y = c_{0} \frac{x^{3}}{6} + c_{1} x + c_{2}$. Now let $y(x) = f(x) + b_{1} x^{3} + b_{2} x^{2} + b_{3} x + b_{4}$ to obtain
\begin{align}
f'' - a f' + f + [b_{1} x^{3} + (b_{2}-3 a b_{1}) x^{2} +(6 b_{1} - 2 a b_{2} + b_{3}) x + (2 b_{2} - a b_{3} + b_{4}) ] = c_{0} \frac{x^{3}}{6} + c_{2}
\end{align}
which leads to $b_{1} = c_{0}/6$, $b_{2} = a c_{0}/2$, $b_{3} = c_{1} - (a^{2} + 1) c_{0}$, $b_{4} = c_{2} + a c_{1} - (2 a + a^{3}) c_{0} $ and 
$$y(x) = f(x) + \frac{c_{0} x^{3}}{6} + \frac{a c_{0} x^{2}}{2} + (c_{1} - (a^{2} + 1) c_{0}) x + c_{2} +a c_{1} -a(2 + a^{2})c_{0}$$. 
The equation for $f(x)$ is $f'' - a f + f = 0$ which has the solution 
\begin{align}
f(x) = e^{ax/2} \left( A e^{\frac{x}{2} \sqrt{a^{2} - 4}} + B e^{- \frac{x}{2} \sqrt{a^{2} - 4}} \right).
\end{align}
Collecting all the parts together leads to
\begin{align}
y(x) = e^{ax/2} \left( A e^{\frac{x}{2} \sqrt{a^{2} - 4}} + B e^{- \frac{x}{2} \sqrt{a^{2} - 4}} \right) + \frac{c_{0} x^{3}}{6} + \frac{a c_{0} x^{2}}{2} + (c_{1} - (a^{2} + 1) c_{0}) x + c_{2} +a c_{1} -a(2 + a^{2})c_{0}.
\end{align}
When $a=2$ this reduces to $y = (A + Bx) e^{x} + \frac{c_{0} x^{3}}{6} + c_{0} x^{2} + (c_{1} - 5 c_{0}) x + c_{2} + 2 c_{1} - 12 c_{0}$. 
A: 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 $$y_p=Ax^3+Bx^2$$
