Let $S=R[x,y,u,v]$ be polynomial ring over noetherian ring $R$.Set $M=S/(xu+yv)$. I guess

$H^2_{(x,y)}(M)=0$ . for example $H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})=0$.

We know that $Supp H^i_I(M)‎\subseteq V(I)\cap Supp(M)$ then $Supp H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=‎\lbrace‎‎ (x,y) \rbrace\cup ‎\lbrace‎‎(x,y,p); \rbrace‎‎$ where $p$ is prime number.

If we could show that for every $P\in V((x,y)), \big(H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})\big)_P=0$(?) it is done.

background: $H^i_I(M)$ means $i$-th local cohomology module of $M$ with respect to ideal $I$. $V(I)=\lbrace P\in Spec(R); I\subseteq P\rbrace $ and $Supp(M)=\lbrace P\in Spec(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$.Furthermore $Supp(R/I)=V(I)$

Edited:

I guess

$$H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$$

We know that if $\operatorname{Supp} H^i_I(M)‎\subseteq V(I)\cap \operatorname{Supp}(M)$, then $$\operatorname{Supp} H^2_{(x,y)}\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=‎\lbrace(x,y) \rbrace\cup ‎\lbrace‎‎(x,y,p) \rbrace‎‎$$ where $p$ is prime number.

If we could show that for every $P\in V((x,y))$, $$\left(H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)\right)_P=0$$

then it is done.

background: $H^i_I(M)$ means $i$-th local cohomology module of $M$ with respect to ideal $I$. $V(I)=\lbrace P\in Spec(R); I\subseteq P\rbrace $ and $\operatorname{Supp}(M)=\lbrace P\in \operatorname{Spec}(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$. Furthermore $\operatorname{Supp}(R/I)=V(I)$.

Let $S=R[x,y,u,v]$ be polynomial ring over noetherian ring $R$.Set $M=S/(xu+yv)$. I guess

$H^2_{(x,y)}(M)=0$ . for example $H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})=0$.

We know that $Supp H^i_I(M)‎\subseteq V(I)\cap Supp(M)$ then $Supp H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=‎\lbrace‎‎ (x,y) \rbrace\cup ‎\lbrace‎‎(x,y,p); \rbrace‎‎$ where $p$ is prime number.

We must show that for every $P\in V((x,y)), \big(H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})\big)_P=0?$.

background: $H^i_I(M)$ means $i$-th local cohomology module of $M$ with respect to ideal $I$. $V(I)=\lbrace P\in Spec(R); I\subseteq P\rbrace $ and $Supp(M)=\lbrace P\in Spec(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$.Furthermore $Supp(R/I)=V(I)$

Let $S=R[x,y,u,v]$ be polynomial ring over noetherian ring $R$.Set $M=S/(xu+yv)$. I guess

$H^2_{(x,y)}(M)=0$ . for example $H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})=0$.

We know that $Supp H^i_I(M)‎\subseteq V(I)\cap Supp(M)$ then $Supp H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=‎\lbrace‎‎ (x,y) \rbrace\cup ‎\lbrace‎‎(x,y,p); \rbrace‎‎$ where $p$ is prime number.

If we could show that for every $P\in V((x,y)), \big(H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})\big)_P=0$(?) it is done.

background: $H^i_I(M)$ means $i$-th local cohomology module of $M$ with respect to ideal $I$. $V(I)=\lbrace P\in Spec(R); I\subseteq P\rbrace $ and $Supp(M)=\lbrace P\in Spec(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$.Furthermore $Supp(R/I)=V(I)$