Understanding the solution to the differential equation $x\dot y=y^{2}-4$ The equation is:
$$x\dot y=y^{2}-4$$
With the starting condition:
$$y(1)=-3$$
So I've done the separation of variables and ended up with:
$$\frac{1}{4}ln\lvert \frac{y-2}{y+2}\rvert =  ln|x|+C$$
Now in the solution, the author has defined $A\ne0$ that's a constant. This is the next step in the solution:
$$\frac{y-2}{y+2}=Ax^4$$
Now they sub in $y=-3$, $x=1$ and they find that $A = 5$. 
The next steps are:
$$ \frac{y-2}{y+2} = 5x^4$$
and the answer is:
$$ y= \frac {10x^{4}+2}{1-5x^4} $$
Can anyone help me understand where the $A$ came from? and why we specifically chose $x$ to the power of $4$?  I'm assuming it's to make solving the starting conditions much easier.
 A: T. Bongers explained your question, and there is a much easier way to solve this problem by assuming that the constant $4$ is $n$.

Let's solve, this first-order nonlinear ordinary differential equation:

*

*$$xy'(x)=y(x)^2-n$$


$$xy'(x)=y(x)^2-n\Longleftrightarrow$$
$$y'(x)=\frac{y(x)^2-n}{x}\Longleftrightarrow$$
$$\frac{y'(x)}{y(x)^2-n}=\frac{1}{x}\Longleftrightarrow$$
$$\int\frac{y'(x)}{y(x)^2-n}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$

To solve this integral, substitute $u=y(x)$ and $\text{d}u=y'(x)\space\text{d}x$:
$$\int\frac{y'(x)}{y(x)^2-n}\space\text{d}x=\int\frac{1}{u^2-n}\space\text{d}u=-\frac{1}{n}\int\frac{1}{1-\frac{u^2}{n}}\space\text{d}u$$
Now, substitute $s=\frac{u}{\sqrt{n}}$ and $\text{d}x=\frac{1}{\sqrt{n}}\space\text{d}u$:
$$-\frac{1}{\sqrt{n}}\int\frac{1}{1-s^2}\space\text{d}s=-\frac{\text{arctanh}\left(s\right)}{\sqrt{n}}+\text{C}=$$
$$-\frac{\text{arctanh}\left(\frac{u}{\sqrt{n}}\right)}{\sqrt{n}}+\text{C}=-\frac{\text{arctanh}\left(\frac{y(x)}{\sqrt{n}}\right)}{\sqrt{n}}+\text{C}$$

$$-\frac{\text{arctanh}\left(\frac{y(x)}{\sqrt{n}}\right)}{\sqrt{n}}=\ln\left|x\right|+\text{C}\Longleftrightarrow$$
$$y(x)=-\sqrt{n}\tanh\left(\sqrt{n}\left(\ln\left|x\right|+\text{C}\right)\right)$$
So, when we know $n=4$:
$$y(x)=-2\tanh\left(2\left(\ln\left|x\right|+\text{C}\right)\right)$$
Now, we can solve $\text{C}$:
$$-3=-2\tanh\left(2\left(\ln\left|1\right|+\text{C}\right)\right)\Longleftrightarrow$$
$$\frac{3}{2}=\tanh\left(2\left(0+\text{C}\right)\right)\Longleftrightarrow$$
$$\text{arctanh}\left(\frac{3}{2}\right)=2\text{C}\Longleftrightarrow$$
$$\frac{\text{arctanh}\left(\frac{3}{2}\right)}{2}=\text{C}$$
So, we get that:
$$y(x)=-2\tanh\left(2\ln\left|x\right|+\text{arctanh}\left(\frac{3}{2}\right)\right)$$
