Find solution set of $\mathbb{C}$-valued equation $y^{(4)}=2y^{(3)}-y''$ Problem:
$\text{Find the solution set of this } \mathbb{C}\text{-valued equation }$
$$y^{(4)}=2y^{(3)}-y''$$
$\text{So we have } y^4=2y^3-y^2 \implies y=1\lor y=0$.
As the solution said, we have $\phi_1(x)=1, \phi_2(x)=x, \phi_3(x)=e^x, \phi_4(x)=xe^x$ and they are linearly independent.
$\mathscr{H}=\{c_1\phi_1+c_2\phi_2+c_3\phi_3+c_4\phi_4|c_1,c_2,c_3,c_4\in\mathbb{C}\}$
I have no idea where those $\phi$ come from, it'll be great, if some one could explain it.
Thanks in advance!
 A: Setting $z(x)=y''(x)$ leads to
$$z''=2z'-z$$
so any solution $z^*(x)$ to this second-order linear equation with constant coefficients leads to a two-parameter family of solutions $y(x)=\iint z^*(x)+Ax+B$, explaining $\phi_1$ and $\phi_2$.
The characteristic polynomial of the second-order equation factors as $(r-1)^2$, leading to $\phi_3(x)=e^x$ and $\phi_4(x)=xe^x$ as is the case when there are repeated roots.
A: $$y^{(4)}=2y^{(3)}-y''$$
$$y^{(4)}-2y^{(3)}+y''=0$$
Rewrite the DE as:
$$(y''e^{-x})''=0$$
and integrate.
A: Let me try to explain it in an easier manner.
We can rewrite the equation as:
$$y^{(4)} - 2y^{(3)} + y^{"} = 0$$
We can express this as a characteristic polynomial to solve for the eigenvalues to the eigenfunction above.
$$ \lambda^4 - 2\lambda^3 + \lambda^2 = 0 $$
$$ \lambda^2(\lambda^2-2\lambda+1)=0 $$
$$ \lambda^2(\lambda-1)^2=0 $$
$$ \lambda_{1,2}=0, \lambda_{3,4}=1$$
We can observe that each root has a algebraic multiplicity of 2, respectively. Alternatively, we can say there are repeated roots.
Hence, the general solution to the ODE above is given by,
$$ y(x) = c_1e^{0x} +c_2xe^{0x} + c_3e^{x} +c_4xe^{x} = c_1 +c_2x + c_3e^{x} +c_4xe^{x} $$ where $c_1, c_2, c_3, c_4$ are constants.
As what you mentioned in your question about $\phi(x)$, the solution to your eigenfunction is represented by the basis $\{1,x, e^x,xe^x\}$ which is linearly independent.
