Brezis' exercise 8.17: the kernel of $A^*$ where $A u=u''-xu'$

Question

Let $I$ be the open interval $(0, 1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Exercise 8.17 Let $H=L^2(I)$ and $A: D(A) \subset H \rightarrow H$ be the unbounded operator defined by $A u=u''-xu'$, whose domain $D(A)$ will be made precise below. Determine $A^{*}, D\left(A^{*}\right), N(A)$, and $N\left(A^{*}\right)$ in the following cases:

1. $D(A)=\left\{u \in H^2 (I) : u(0)=u(1)=0\right\}$.
2. $D(A)=H^2(I)$.
3. $D(A)=\left\{u \in H^2(I) : u(0)=u(1)=u'(0)=u'(1)=0\right\}$.
4. $D(A)=\left\{u \in H^2(I) : u(0)=u(1)\right\}$.

There are possibly subtle mistakes that I could not recognize in my below attempt. Could you please have a check on my computation of $N(A^*)$?

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In a previous thread, I proved that $D(A^*) \subset H^2 (I)$ and that $A^*v = v''+xv'+v = (v'+xv)'$. With each boundary condition, I determined $D(A^*)$ as follows.

1. $D(A^*) = \{ v \in H^2 (I) : v(1) = v(0) = 0 \}$.

2. $D(A^*) = \{ v \in H^2 (I) : v(1) = v(0) = v'(1) = v'(0) = 0 \}$

3. $D(A^*) = H^2 (I)$.

4. $D(A^*) = \{ v \in H^2 (I) : v(1) = v(0) = v'(1) - v'(0) = 0 \}$.

We have $N(A^*) = \{v \in D(A^*) : v'+xv \text{ is constant}\}$. With each boundary condition, I determined $N(A^*)$ as follows.

1. $N(A^*)$ consists of those $v \in H^2 (I)$ such that there is a constant $C$ such that $v'+xv=C$ and $v(1) = v(0) = 0$.

2. $N(A^*)$ consists of those $v \in H^2 (I)$ such that $v'+xv=0$ and $v(1) = v(0) = 0$.

3. $N(A^*)$ consists of those $v \in H^2 (I)$ such that there is a constant $C$ such that $v'+xv=C$.

4. $N(A^*)$ consists of those $v \in H^2 (I)$ such that there is a constant $C$ such that $v'+xv=C$ and $v(1) = v(0) = 0$.