Cooking up a $1$-form satisfying two equalities 
Denote the standard coordinates on $\mathbb{R}^2$ by $x,y$ and let $$X = -y\frac{\partial}{\partial x} + x \frac{\partial}{\partial y} \ \text{ and } Y = x\frac{\partial}{\partial x} + y \frac{\partial}{\partial y}.$$ be vector fields on $\mathbb{R}^2$. Find a $1$-form $\omega$ on $\mathbb{R}^2 \setminus \{(0,0)\}$ such that $\omega(X)=1$ and $\omega(Y)=0$.

If we let $\omega = adx +bdy$, then computing $\omega(X)$ I concluded that $$\omega(X)= -ay+bx$$ and likewise $$\omega(Y)  =ax+ by.$$
Now I'm looking for $a$ and $y$ such that $$\begin{align*}-ay+bx&= 1 \\ ax+by &= 0. \end{align*}$$
Is there a way for me to cook up such an $a$ and $b$ just from these two equations or do I need to do something clever here? Adding these two I got that $$a(x-y)+b(x+y)=1$$so any $a$ and $b$ satisfying this would also work.
 A: This is a system of linear equations in variables $a$ and $b$, where the coefficients happen to be functions of $x$ and $y$, but you can still use any technique to solve the system. For example, with matrices, the system is equivalent to
$$
\begin{bmatrix} 
-y & x \\
 x & y
\end{bmatrix}
\begin{bmatrix} 
a \\ b
\end{bmatrix}
= 
\begin{bmatrix} 
1 \\ 0
\end{bmatrix}. 
$$
Try to work this out yourself before revealing the spoilers:

 \begin{align} \begin{bmatrix} a \\ b \end{bmatrix} &= \begin{bmatrix} -y & x \\[2pt]  x & y \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\[2pt] &= -\frac{1}{x^2 + y^2} \begin{bmatrix}  y & -x \\ -x & -y \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\[2pt] &= \frac{1}{x^2 + y^2} \begin{bmatrix} -y & x \\ x & y \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\[2pt] &= \frac{1}{x^2 + y^2} \begin{bmatrix} -y \\  x \end{bmatrix}, \end{align}

i.e. the $1$-form is

 $$ \omega = \frac{-y \, dx + x \, dy}{x^2 + y^2} $$.

Notice that it happens that this form is well-defined (and smooth) on the entire plane other than the origin! It didn't have to turn out that way, but it did, so we're done.
It is in some sense dual to the rotation field of constant angular frequency, hence it evaluates on such a field to give the value $1$ and is orthogonal to the radial field, giving value $0$.
