Factorize the differential operator $[D^2-(x^2+1)]$ into two first order linear operators

Question

I am Trying to factorize the differential operator $[D^2-(x^2+1)]$ into two first order linear operators. But I have not been able to.

I am trying to split it into $[D+A][D-B]f=[D^2-(x^2+1)]f$. But at the end I get a really ugly differential equation that I cannot solve (It should be an easy one).

Is there any hint?

Thanks

@PeterTamaroff: What I have seen in operator $D$ is $D$ needs to act on a function as $Dy$ not $D\pm y$. Don't you think it is a typo in the question? — Mikasa, Mar 10 '13 at 18:30
@BabakS. No. It's OK as it stands. One usually thinks about operators such as $x\frac{d}{dx}$ or $\frac{d}{dx}-x$, which are useful. — Pedro, Mar 10 '13 at 18:36
it is [D+A][D-B] Where A, B are functions. The operator is acting in f. — user66130, Mar 10 '13 at 18:37
Hint: what is $D( x \phi(x)) - x D(\phi(x))$ for any function $\phi(x)$? — achille hui, Mar 10 '13 at 18:37
@Peter Difference of squares formula requires commutativity. But $\rm: D f !-! f D = f' = 0 \iff f,$ is constant, i.e as linear operators, the derivative $\rm,D,$ commutes only with constant functions. — Math Gems, Mar 10 '13 at 18:41
@MathGems Hehehe, I'd like to ask you a little about Jacobson. — Pedro, Mar 10 '13 at 18:49
@MathGems Oh, I don't really have a question, but'd like to discuss stuff. I'll delete the comments. — Pedro, Mar 10 '13 at 18:58

score 3 · Accepted Answer · answered Mar 10 '13 at 19:08

Factorization means, that for any differentiable function $f(x)$, and for some $a(x)$ and $b(x)$ we have: $$ (D^2 - (x^2+1)) \circ f(x) = f^{\prime\prime}(x) - (x^2+1) f(x) = \left( D + a(x) \right) \left(D + b(x) \right) \circ f(x) $$ Start expanding the right-hand-side: $$\begin{eqnarray} \left( D + a(x) \right) \left(D + b(x) \right) \circ f(x) &=& \left( D + a(x) \right) \circ \left( f^{\prime}(x) + b(x) f(x)\right) \\ &=& f^{\prime\prime}(x) + \left(b(x) f(x)\right)^\prime + a(x) f^\prime(x) + a(x) b(x) f(x) \\ &=& f^{\prime\prime}(x) + \left( a(x) + b(x) \right) f^\prime(x) + \left( a(x) b(x) + b^\prime(x) \right) f(x) \end{eqnarray} $$ Since the identity must hold for arbitrary $f(x)$ we should have $a(x) + b(x) = 0$ and $a(x) b(x) + b^\prime(x) = -x^2-1$, giving $a(x) = -b(x) = x$. Thus $$ (D^2 - (x^2+1)) = (D + x)( D-x) $$

thanks. The las differential equation how do you solve it without inspection? — user66130, Mar 10 '13 at 19:35
@Sasha What are the conditions to find a factorisation for any linear differential operator? thanks for your time — schzan, Dec 18 '14 at 14:44

Factorize the differential operator $[D^2-(x^2+1)]$ into two first order linear operators

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