Why can linear independence of $e^{ax +by^2}$ be proven without considering $y$? Linear independence of $e^{at}$ has been answered multiple times. My favorite one is by Marc van Leeuwen in this one: Proof of linear independence of $e^{at}$. The answer uses the property that $e^{at}$ are eigenfunctions of the differentation operation. Now if we instead have an exponential function in two variables, say $e^{ax + by^2}$, it seems to me the linear independence of these functions can too be proved with the same technique, except now using the partial derivative operator:
Proof by induction as in the link except here $e^{a_1x + b_1y^2}$, $e^{a_2x + b_2y^2}$, $...$,$e^{a_{n-1}x + b_{n-1}y^2}$ are assumed linearly independent. As in Marc's proof, we then assume that $e^{a_1x + b_1y^2}$, $e^{a_2x + b_2y^2}$, $...$,$e^{a_{n}x + b_{n}y^2}$ are in turn dependent and thus have:
$e^{a_{n}x + b_{n}y^2}= c_1e^{a_1x + b_1y^2} +  c_2e^{a_2x + b_2y^2} + ... + c_{n-1}e^{a_{n-1}x + b_{n-1}y^2}$. Applying operator $\frac{\partial}{\partial x} - a_nI$ will give  $0= c_1(a_1-a_n)e^{a_1x + b_1y^2} +  c_2(a_2-a_n)e^{a_2x + b_2y^2} + ... + c_{n-1}(a_{n-1}-a_n)e^{a_{n-1}x + b_{n-1}y^2}$ which would thus require all c to be zero, which essentially completes the induction proof (other argumentation as per link).
My question is: why does $y^2$ not seem to have any effect on the linear independence. Where does this stem from?
edit: assume all $a_k$ and $b_k$ are distinct.
 A: Your proof essentially repeats the proof that the functions $e^{a_j x}$ are linearly independent. The “$y$-terms” have “no effect” because they occur as (non-zero) factors $e^{b_j y^2}$ which are constant with respect to $x$.
What you observed is this: If $g_1, \ldots, g_n : A \to \Bbb R$ and $h_1, \ldots, h_n : B \to \Bbb R$ are functions such that


*

*$g_1, \ldots, g_n$ are linearly independent, and

*there is a $y_0 \in B$ such that $h_j(y_0) \ne 0$ for all $j$,


then the functions $f_j : A \times B \to \Bbb R$, $f_j(x, y) = g_j(x) h_j(y)$, $j=1, \ldots, n$, are linearly independent.
The proof is straight forward: If $c_1, \ldots, c_n \in \Bbb R$ with
$$
 \sum_{j=1}^n c_j g_j(x) h_j(y) = 0 \text{ for all } (x, y) \in A \times B
$$
then in particular
$$
 \sum_{j=1}^n \bigl( c_j h_j(y_0) \bigr) g_j(x) = 0 \text{ for all } x \in A 
$$
Since the $g_j$ are linearly independent it follows that 
$$
c_j h_j(y_0) = 0\text{ for  } j = 1, \ldots, n 
\implies c_j = 0  \text{ for  } j = 1, \ldots, n \,.
$$
In your case $g_j(x) = e^{a_j x}$ and $h_j(y) = e^{b_j y^2}$.
