$-f^{(5)}+2f^{(3)}-f'=0$ Solve the following equation $$-f^{(5)}+2f^{(3)}-f'=0$$
My idea
The equation resembles the polynomial $-x^5+2x^3-x$ which can be written as $(x^2+x)(x-1)(-x^2+1)$. Now taking $«x^2=f''(x)»,\ldots $, we have to solve
$(f''+f')(f'-f)(-f''+f)=0$.
As we know (from Lagrange's theorem) the solutions of each of these three equations, this leads to the system$$\left [\begin{matrix}1&1&0\\0&1&-1\\-1&0&1\end{matrix}\right |\left .\begin{matrix}c_1\cos x+c_2\sin x\\c_3e^x\\c_4e^x+c_5e^{-x}\end{matrix}\right ].$$And how to solve the system is reasonably simple, but I don't see the solution presented by Wolfram Alpha.
 A: Method 1:  Characteristic eqn.
$$x = e^{rt}.$$
$$-r^5e^{rt}+2r^3e^{rt}-re^{rt} = 0.$$
$$(-r^5+2r^3-r) = 0.$$
$$r(r^4-2r^2+1) = 0\implies r(r^2-1)(r^2-1) = 0.$$
$$r = 0,\quad r = -1\text{ (double root)},\quad r = 1\text{ (double root)}.$$
General solution is therefore: $$x(t) = c_0+(c_1+c_2t)e^t+(c_3+c_4t)e^{-t}.$$
Method 2:  Factoring:
$$\frac{d}{dt}\left(\frac{d^2}{dt^2}-1\right)\left(\frac{d^2}{dt^2}-1\right)f = 0.$$
Integrate once to get
$$\left(\frac{d^2}{dt^2}-1\right)\left(\frac{d^2}{dt^2}-1\right)f = c_0.$$
Let $g(t) = f''-f$. Then,
$$\left(\frac{d^2}{dt^2}-1\right)g = c_0.$$
$$g(t) = -c_0+c_1e^t+c_2e^{-t}.$$
Then,
$$\left(\frac{d^2}{dt^2}-1\right)f = -c_0+c_1e^t+c_2e^{-t}.$$
Solve this using method of characteristics, and you find same solution as before.
This is a good question, because it shows us that factoring differential operators cannot be viewed in exactly the same way as factoring polynomials.  As the OP demonstrates, we can also factor the operator as
$$\left(\frac{d^2}{dt^2}+\frac{d}{dt}\right)\left(\frac{d}{dt}-1\right)\left(-\frac{d^2}{dt^2}+1\right)f = 0.$$
Unlike for polynomials, we cannot set each factor equal to zero to obtain the correct solution.  From the third factor, we would get as a part of the solution basis sines and cosines, which is not what we obtained above.
