I've been looking through old questions for some time and I found an interesting topic (Complex differential equation). I decided to try solving a complex differential equation with a similar premise. I looked at the equation $$z' = \overline{\mathbb{z}} +it$$ I followed a similar strategy to the post linked, giving $z'' = \overline{\mathbb{z'}} + i$ and then taking the conjugate of the original equation, where $\overline{\mathbb{z'}} = z - it$
Finally, I plugged this back into the equation for $z''$ to get $$z'' = z + i -it$$ $$z'' - z = i - it$$
This gives a non-homogeneous differential equation without any conjugates, which I am more familiar with solving. I decided to attempt solving it through the use of power series, where the solution, $z(t)$ is of the form $$z(t) = \displaystyle\sum_{n=0}^{\infty} C_nt^n$$
I then took the second derivative of this and plugged it into our derived formula to get $$\displaystyle\sum_{n=0}^{\infty} C_n(n)(n-1)t^{n-2} - \displaystyle\sum_{n=0}^{\infty} C_nt^n = i - it$$
Fixing the power and index of term 1 then yields $$\displaystyle\sum_{n=0}^{\infty} t^n[C_{n+2}(n+2)(n+1) - C_n] = i - it $$
Because the right hand side is non-zero, we obviously can't assume that the terms in square brackets sum to zero for all n. From here I decided to evaluate the summation at the first few values of n and compare it to the right hand side to build a recursive relation.
$n =0$ $$t^0(C_2(2)(1) - C_0) = t^0(i)$$ $$C_2(2)(1) - C_0 = i$$ $$C_2 = \frac{C_0 + i}{2}$$
$n =1$ $$t^1(C_3(3)(2) - C_1) = t^1(-i)$$ $$C_3 = \frac{C_1-i}{(3)(2)}$$
$n=2$ $$t^2(C_4(4)(3) - C_2) = t^2(0)$$ $$C_4 = \frac{C_0+i}{(4)(3)(2)}$$
$n=3$ $$t^3(C_5(5)(4) - c_3) = t^3(0)$$ $$C_5 = \frac{C_1 - i}{(5)(4)(3)(2)}$$
This pattern continues for later values of n, giving us the relations
$$ C_n = \frac{C_0+i}{n!} , n \in \mathbb{2Z} , n\geq2$$ $$ C_n = \frac{C_1-i}{n!} , n \in \mathbb{2Z+1} , n>2$$
With these relations, I was able to then build the final solution of $z(t)$ $$z(t) = \displaystyle\sum_{n=0}^{\infty} C_nt^n$$ $$z(t) = C_0+C_1t+(C_0+i)\displaystyle\sum_{n=1}^{\infty} \frac{t^{2n}}{(2n)!}+(C_1-i)\displaystyle\sum_{n=1}^{\infty} \frac{t^{2n+1}}{(2n+1)!}$$
We can then use the hyperbolic sine and cosine taylor series to simplify our answer, where $$\cosh(t) = \displaystyle\sum_{n=0}^{\infty} \frac{t^{2n}}{(2n)!} = \displaystyle\sum_{n=1}^{\infty} \frac{t^{2n}}{(2n)!}+1$$ $$\sinh(t) = \displaystyle\sum_{n=0}^{\infty} \frac{t^{2n+1}}{(2n+1)!} = \displaystyle\sum_{n=1}^{\infty} \frac{t^{2n+1}}{(2n+1)!}+t$$
This gave me my final solution for $z(t)$ $$z(t) = (C_0+i)\cosh(t) + (C_1-i)\sinh(t)-i-it$$
Finally, my problem lies in actually plugging the solution back into the original differential equation. I won't write it out now because I've definitely written too much already, but it's clear that both sides would not equal each other. I was hoping for someone to know where I went wrong, or if there's a better way to solve this type of problem. Sorry for any weird formatting or notation. This is my first time using stack exchange and latex. I've also only learned math up to Calc 3 and ODEs, so there might be some concepts I'm missing that screwed my answer. Thanks!
