# I am trying to solve a first order differential equation with the condition that g ( y

I am trying to solve a first order differential equation with the condition that $g\left(y\right)=0$ if $y=0$:
$\begin{array}{rl}& a{g}^{\prime }\left(cy\right)+b{g}^{\prime }\left(ey\right)=\alpha \\ \text{(1)}& & g\left(0\right)=0,\end{array}$
where parameters a,b,c,e are real nonzero constants; $\alpha$ is a complex constant; function $g\left(y\right):\mathbb{R}\to \mathbb{C}$ is a function mapping from real number y to a complex number. The goal is to solve for function $g\left(\cdot \right)$. This is what I have done. Solve this differential equation by integrating with respect to y:
$\begin{array}{r}\frac{a}{c}g\left(cy\right)+\frac{b}{e}g\left(ey\right)=\alpha y+\beta ,\end{array}$
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
$\begin{array}{r}\frac{a}{c}g\left(cy\right)+\frac{b}{e}g\left(ey\right)=\alpha y.\end{array}$
The background of this problem is Cauchy functional equation, so my conjecture is one solution could be $g\left(y\right)=\gamma y$. Plugging in $g\left(y\right)=\gamma y$, I get $\gamma =\frac{\alpha }{a+b}$, which implies that one solution is $g\left(y\right)=\frac{\alpha }{a+b}y$. Then, I move on to show uniqueness. I define a vector-valued function $h\equiv \left({h}_{1},{h}_{2}{\right)}^{T}$ such that
$\begin{array}{rl}{h}_{1}\left(y\right)& =\frac{a}{c}{g}_{1}\left(cy\right)+\frac{b}{e}{g}_{1}\left(ey\right)\\ {h}_{2}\left(y\right)& =\frac{a}{c}{g}_{2}\left(cy\right)+\frac{b}{e}{g}_{2}\left(ey\right),\end{array}$
where $g\left(y\right)\equiv {g}_{1}\left(y\right)+i{g}_{2}\left(y\right)$. Then, I rewrite this differential equation as
$\begin{array}{rl}& {h}^{\prime }\left(y\right)=\alpha \\ \text{(2)}& & h\left(0\right)=0,\end{array}$
where $\alpha \equiv \left({\alpha }_{1},i{\alpha }_{2}{\right)}^{T}$. 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 ${g}_{1}\left(0\right)=0,{g}_{2}\left(0\right)=0$ 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.
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jugf5
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.
$\begin{array}{r}a{g}^{\prime }\left(cy\right)+b{g}^{\prime }\left(ey\right)=\alpha \end{array}$
Define $f\left(\cdot \right)\equiv {g}^{\prime }\left(\cdot \right)$:
$\begin{array}{r}af\left(cy\right)+bf\left(ey\right)=\alpha \end{array}$
View this equation as an infinite system of algebraic equations in unknowns (a,b). For example,
$\begin{array}{rl}af\left(0\right)+bf\left(0\right)& =\alpha \\ af\left(c\right)+bf\left(e\right)& =\alpha \\ af\left(2c\right)+bf\left(2e\right)& =\alpha \\ & \dots \end{array}$
Since a and b are non-zero, this system of equations is solvable if and only if function $f\left(\cdot \right)$ is a constant function. Thus, function ${g}^{\prime }\left(\cdot \right)$ is a constant function which further implies that function $g\left(y\right)=\beta y$ where $\beta$ is a complex constant.