# Thread: Exact LDE factorizing

1. ## Exact LDE factorizing

Hi all,
I'm currently working on a question that I can't fully work out. Can someone look over my working and see where I may have possibly gone wrong and maybe point me in the right direction?
The question is,
Given a LDE $F=b_{0}y''+b_{1}y'+b_{2}y=0$ and the relationship $b_{0}''-b_{1}'+b_{2}=0$, show that the factorization of the LDE is $F=D[b_{0}D+(b_{1}-b_{0}')]y$.

From the relationship, I can use $b_{2}=b_{1}'-b_{0}''$ and I can factorize that into $b_{2}=D[b_{1}-b_{0}']$.
Factorizing: F = $[b_{0}D^2+ b_{1}D+b_{0}]y$

Therefore, $F= D[b_{0}D+ b_{1}+(b_{1}-b_{0})]y$.
I have that $b_{1}$ that I'm not sure how to get rid of.
Thank you for your time.

2. Expand $F$:

$F=D[b_0 D+(b_1-b_0^{'})]y$

$F=D[b_0y'+(b_1-b_0^{'})y]$

$F=b_0 y''+y'b_0^{'}+(b_1-b_0^{'})y'+y(b_1^{'}-b_0^{''})$

now let $b_2=b_1^{'}-b_0^{''}$

3. Thank you, I never thought about working backwards.