Solve the following differential equation in $\mathbb{C}$. $$x' = \frac{cx}{|x|}$$
Here's the original question I encountered: There are A, B, C and D sitting on 4 vertexes of a square. They are all moving at a constant speed of $v$ towards others- A towards B, B towards C, C towards D, and D towards A. Since this movement is entirely symmetric(let the origin be in the middle of the square), $$|OA| = |OB|$$ OAB makes an isosceles right triangle, and hence $\angle{OAB} = \frac{\pi}{4}$; thus
$$ x'_A = \left(-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i \right)\frac{x_A}{|x_A|}$$ Or in other words, $x' = A\frac{x}{|v|}$ with $A = [a, b; c, d]$ ,$a=d$ and $b=-c$.
I have solved this problem by letting $x = r\psi$ with $|\psi|=1$, which gave me the solution-something in the form of $r=r_0 - ut$. But I would like to ask if there is a more general way of solving ODE like this- perhaps without doing $x = r \psi$? Thanks in advance.
Just directly integrating:
Suppose for a real constant $c\in\mathbb{R}$ and an unknown function whose $Im(x(t))\subseteq\mathbb{R}$, we have: $$x' = \frac{cx}{|x|},$$ where $|x(t)| = \bigg\{\begin{matrix}x(t)&&\text{ if }x(t)>0\\-x(t)&&\text{ if }x(t)<0\end{matrix}$.
Then provided $\color{red}{x\neq 0}$, manipulating and integrating w.r.t. the curve parameter gives: $$\int\frac{|x|}{x}x'dt = \int cdt,$$ which upon integrating the LHS by parts with $u:= |x|$ and $dv:= \frac{x'}{x}dt$ gives:
Recall: IBP $\color{blue}{\bigg[\large\int \normalsize u\text{ }dv = u\text{ }v-\large\int \normalsize v\text{ }du\bigg]}$
$$LHS = |x|\big(\int\frac{x'}{x}dt\big)- \int\bigg(\big(\int\frac{x'}{x}dt\big)\cdot |x'|\bigg)dt.$$ To continue, we must evaluate the inner integral: Apply IBP again with $u:= \frac{1}{x}$ and $dv:= x'dt$ $$\int\frac{x'}{x}dt = \frac{1}{x}\cdot x-\int x\cdot\frac{-1}{x^2}dt = 1 + \int \frac{1}{x}dt = 1+ln|x|+C$$ So we now have: $$LHS = |x|\cdot(ln|x|+C)-\int(ln|x|+C)\cdot|x'|dt,$$ where $C$ absorbed the $1$. This, by Linearity of Integrals, equals: $$=|x|(ln|x|+C)-\int|x'|ln|x|dt-\int C|x'|dt$$ $$= |x|ln|x|-\int|x'|ln|x|dt.$$
Using IBP a third time with $u:= ln|x|$ and $dv:= |x'|dt$ we get: $$=|x|ln|x| - \bigg[ln|x|\cdot(|x|+D) - \int(|x|+D)\cdot\frac{1}{|x|}x'dt\bigg]$$ $$= |x|ln|x|-\bigg[|x|ln|x|+Dln|x| - \int x'dt-D\int \frac{x'}{|x|}dt\bigg].$$ Then distributing the negative, canceling the $|x|ln|x|$, and choosing $D=0$, we have: $$LHS = \int x'dt = x+E$$ So finally, connecting this with the RHS ($\int cdt$), we get the original ODE implies: $$x+E = ct+F$$
$$\implies x = ct+a\text{ }\text{ for }\text{ }c,a\in\mathbb{R}$$
(by absorbing $-E$ into $F$ and renaming $F$ as $a$).
In a long winded proof, we found "a" potential solution to the ODE $$x' = \frac{cx}{|x|}$$ is the line:
$$\color{blue}{x = ct+a}.$$
Now we just test it: $$x' = c\text{ }\text{ and }\text{ }\frac{cx}{|x|} = \frac{c(ct+a)}{|ct+a|} = \bigg\{\begin{matrix}c&&\text{ if }ct+a > 0 \\ -c&&\text{ if }ct+a < 0\end{matrix}.$$ So we see the requirement for our solution to work is just: $$ct+a > 0\text{ }\text{ OR }\text{ }\color{red}{t>-\frac{a}{c}}.$$
Note that potentially more solutions may be found by not setting $D=0$.