What I have so far: I really appreciate any help.. I am stuck on this.
I want to get
We want to determine the 3rd order 3-step BDF method $$\sum_{j=0}^r a_jU^{n+j}=kB_rf^{n+r} \Rightarrow a_0U^n+a_1U^{n+1}+a_2U^{n+2}+a_3B^{n+3}=kB_3f(U^{n+3})$$
We have that $$u'(t_{n+3})=\frac{1}{kB_3} \big [a_0U^n +a_1U^{n+1}+a_2U^{n+2}+a_3B^{n+3}\big ]$$\
We want to use a Taylor Series expansion about the point $t_{n+3}$: $$U_{n+3}=U_{n+3}\tag{1}$$ Knowing that $t_{n+2}=t_{n+3}-k$, we have that
\begin{align*} U_{n+2}&=U_{n+3}+U'_{n+3}(t_{n+3}-k-t_{n+3})+\frac{1}{2}U''_{n+3}(t_{n+3}-k-t_{n+3})^2\\&\quad+\frac{1}{6}U'''_{n+3}(t_{n+3}-k-t_{n+3})^3+O(k^4)\\ &=U_{n+3}-kU'_{n+3}+\frac{k^2}{2}U''_{n+3}-\frac{k^3}{6}U'''_{n+3}+O(k^4)\tag{2} \end{align*} Similarly, we can find $U_{n+1}$ and $U_{n}$.
\begin{align*} U_{n+1}&=U_{n+3}+U'_{n+3}(t_{n+3}-2k-t_{n+3})+\frac{1}{2}U''_{n+3}(t_{n+3}-2k-t_{n+3})^2\\ &\quad+\frac{1}{6}U'''_{n+3}(t_{n+3}-2k-t_{n+3})^3+O(k^4)\\ &=U_{n+3}-2kU'_{n+3}+2k^2U''_{n+3}-\frac{8k^3}{6}U'''_{n+3}+O(k^4)\tag{3} \end{align*} \begin{align*} U_{n}&=U_{n+3}+U'_{n+3}(t_{n+3}-3k-t_{n+3})+\frac{1}{2}U''_{n+3}(t_{n+3}-3k-t_{n+3})^2\\ &\quad+\frac{1}{6}U'''_{n+3}(t_{n+3}-3k-t_{n+3})^3+O(k^4)\\ &=U_{n+3}-3kU'_{n+3}+\frac{9k^2}{2}U''_{n+3}-\frac{27k^3}{6}U'''_{n+3}+O(k^4)\tag{4} \end{align*} Then we have \begin{align*} u'(t_{n+3})= \frac{1}{kB_3}\bigg[a_0\big[U_{n+3}-3kU'_{n+3}+\frac{9k^2}{2}U''_{n+3}-\frac{27k^3}{6}U'''_{n+3} \big] + a_1 \big[U_{n+3}-2kU'_{n+3}+2k^2U''_{n+3} \\ -\frac{8k^3}{6}U'''_{n+3} \big] +a_2 \big[U_{n+3}-kU'_{n+3}+\frac{k^2}{2}U''_{n+3}-\frac{k^3}{6}U'''_{n+3} \big]+a_3\big[U_{n+2} \big]\bigg] \end{align*}
I think I need to solve a system of some sort. I tried a matrix like this where a_n is a0. a_n+1 is a1, etc..
by following this post, How do you derive the backward differentiation formula of 3rd order using interpolating polynomials? but it didn't produce the correct solution :/
Thanks.
Your equations that I have taken the liberty of numbering $(1)$ through $(4)$ read, as a matrix equation $$\begin{bmatrix}1&0&0&0\\ 1&-1&\frac12&-\frac16\\ 1&-2&2&-\frac43\\ 1&-3&\frac92&-\frac92\end{bmatrix}\begin{bmatrix}U_{n+3}\\kU_{n+3}^{\prime}\\k^2U_{n+3}^{\prime\prime}\\k^3U_{n+3}^{\prime\prime\prime}\end{bmatrix}=\begin{bmatrix}U_{n+3}\\U_{n+2}\\U_{n+1}\\U_n\end{bmatrix}$$ We can solve this for $$\begin{align}kU_{n+3}^{\prime}&=\begin{bmatrix}0&1&0&0\end{bmatrix}\begin{bmatrix}1&0&0&0\\ 1&-1&\frac12&-\frac16\\ 1&-2&2&-\frac43\\ 1&-3&\frac92&-\frac92\end{bmatrix}^{-1}\begin{bmatrix}U_{n+3}\\U_{n+2}\\U_{n+1}\\U_n\end{bmatrix}\\ &=\frac{11}6U_{n+3}-3U_{n+2}+\frac32U_{n+1}-\frac13U_n\end{align}$$ And that works out to $$U_{n+3}-\frac{18}{11}U_{n+2}+\frac9{11}U_{n+1}-\frac2{11}U_n=\frac6{11}kU_{n+3}^{\prime}$$ As hoped for.
EDIT: Looking at your final matrix equation you should have gotten the same formula as I did. How did you try to solve it? There is a typo in that expression $a_3[U_{n+2}]$ but it didn't propagate into your matrix equation.