But can $y'=y^i$ where $i:=\sqrt{-1}$?

Question

So I was doing some math for fun (because why not), and started to do some first order nonlinear ODEs. After a while, I came up with this:$$y'=y^{i}$$where $i=\sqrt{-1}$, which I thought that I might be able to do. Here is my attempt at doing so:$$\begin{align}y'=y^{i}\implies&\,\dfrac{dy}{dx}=y^{i}\implies\dfrac{dy}{y^{i}}=dx\\\implies&\,\int\dfrac{dy}{y^{i}}=\int dx\\\implies&\,\dfrac{1+i}{2y^{1-i}}+c_0=x+c_1\\\implies&\,\dfrac1{y^{1-i}}=\dfrac{2x+c_2}{1+i}\implies y^{1-i}=\dfrac{1+i}{2x+c_2}\\\implies&\,y(x)=\sqrt[1-i]{\dfrac{1+i}{2x+c_2}},x\ne\dfrac{c_2}2\text{ or }\sqrt{\left(\dfrac{1+i}{2x+c_2}\right)^{1+i}},x\ne\dfrac{c_2}2\end{align}$$because generally$$\dfrac1z\gets\dfrac1z\left(\dfrac{\overline z}{\overline z}\right)=\dfrac{\overline z}{z\overline z}\text{ where }z=a+bi\text{ with }a,b\in\mathbb R$$where we have that $\overline z$ is the conjugate of our complex number $z$ which implies that$$\sqrt[z]{f(x)}=(f(x))^{1/z}\gets\sqrt[z\overline z]{(f(x))^{\overline z}}$$and we cannot have $x=\frac{c_2}2$ here because the solution to any $ax+b=0$ is $x=-b/a$ with $a\ne0$ because division by zero is not possible. So my question is:

Is my solution to this first order nonlinear ODE correct, or what would I do to get the correct solution?

Answer 1

One of the solutions should be $$y=(1-i)^{\frac{1+i}{2}}\, (x+c_1){}^{\frac{1+i}{2}}$$

$$(1-i)^{\frac{1+i}{2}}=\sqrt[4]{2 e^{\pi /2}}\left(\cos \left(\frac{\pi }{8}-\frac{\log (2)}{4}\right)-i \sin \left(\frac{\pi }{8}-\frac{\log (2)}{4}\right) \right)$$

However, with the complex numbers, we never really know! There may be an infinite amount of solutions.