Solve a first order differential equation with parameters inside functions: $ag'(cy)+bg'(ey)=\alpha$

Question

I am trying to solve a first order differential equation with the condition that $g(y)=0$ iff $y=0$:
\begin{align*}
&ag'(cy)+bg'(ey)=\alpha\\
&g(0)=0, \tag{1}
\end{align*}
where parameters $a,b,c,e$ are real nonzero constants; $\alpha$ is a complex constant; function $g(y):\mathbb{R}\to \mathbb{C}$ is a function mapping from real number $y$ to a complex number. The goal is to solve for function $g(\cdot)$. This is what I have done. Solve this differential equation by integrating with respect to $y$:
\begin{align*}
\frac{a}{c}g(cy)+\frac{b}{e}g(ey)=\alpha y+\beta,
\end{align*}
where $\beta$ is another complex constant. Plugging in $y=0$ and using the fact that $g(0)=0$, we have $\beta=0$. Therefore, we have
\begin{align*}
\frac{a}{c}g(cy)+\frac{b}{e}g(ey)=\alpha y.
\end{align*}
The background of this problem is Cauchy functional equation, so my conjecture is one solution could be $g(y)=\gamma y$. Plugging in $g(y)=\gamma y$, I get $\gamma=\frac{\alpha}{a+b}$, which implies that one solution is $g(y)=\frac{\alpha}{a+b} y$. Then, I move on to show uniqueness. I define a vector-valued function $h\equiv (h_1,h_2)^T$ such that
\begin{align*}
h_1(y)&=\frac{a}{c}g_1(cy)+\frac{b}{e}g_1(ey)\\
h_2(y)&=\frac{a}{c}g_2(cy)+\frac{b}{e}g_2(ey),
\end{align*}
where $g(y)\equiv g_1(y)+ig_2(y)$. Then, I rewrite this differential equation as
\begin{align*}
&h'(y)=\alpha\\
&h(0)=0, \tag{2}
\end{align*}
where $\alpha\equiv (\alpha_1,i\alpha_2)^T$. By the uniqueness theorem of first order differential equation, solution $h(y)$ is unique. I have two questions. First, I think equation (1) and (2) should be equivalent. However, it seems that equation (1) can imply equation (2) but equation (2) may not imply equation (1). This is because $h(0)=0$ may imply either $g_1(0)=0,g_2(0)=0$ or $\frac{a}{c}+\frac{b}{e}=0$. Second, I have only proved that $h(y)$ is unique. How should I proceed to show $g(y)$ is also unique.

Answer 1

For people who are interested in my question, I got an answer from a friend. This is a smart way to show uniqueness of solution. \begin{align*} ag'(cy)+bg'(ey)=\alpha \end{align*} Define $f(\cdot)\equiv g'(\cdot)$: \begin{align*} af(cy)+bf(ey)=\alpha \end{align*} View this equation as an infinite system of algebraic equations in unknowns $(a,b)$. For example, \begin{align*} af(0)+bf(0)&=\alpha\\ af(c)+bf(e)&=\alpha\\ af(2c)+bf(2e)&=\alpha\\ &\dots \end{align*} Since $a$ and $b$ are non-zero, this system of equations is solvable if and only if function $f(\cdot)$ is a constant function. Thus, function $g'(\cdot)$ is a constant function which further implies that function $g(y)=\beta y$ where $\beta$ is a complex constant.