A specific homogeneous polar differential equation In an assignment of our school, we are asked to solve
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + y - 3}{x - y - 1}$$
by turning it into a homogeneous polar differential equation (equation of the form $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = F\left(\frac{y}{x}\right)$) using substitutions $x = X + a$, $y = Y + b$. My solution was:
Firstly, I determined substitutions $x = X + 2$, $y = Y + 1$, such that
$$\frac{\mathrm{d}Y}{\mathrm{d}X} = \frac{X + Y}{X - Y}$$
Then, let $Y = Xv$, thus
$$\begin{aligned}
v + X\frac{\mathrm{d}v}{\mathrm{d}X} &= \frac{X + Xv}{X - Xv} \\
\int \frac{1 - v}{1 + v^2}\ \mathrm{d}v &= \int \frac{\mathrm{d}X}{X}
\end{aligned}$$
The left-hand side, specifically, gives
$$\int \frac{\mathrm{d}v}{1 + v^2} - \int \frac{v}{1 + v^2}\ \mathrm{d}v = \tan^{-1} v - \frac{\ln \left(1 + v^2\right)}{2} + \mathrm{constant}$$
Therefore,
$$\tan^{-1} v - \frac{\ln \left(1 + v^2\right)}{2} = \ln \left|X\right| + \mathrm{constant}$$
i.e.
$$2\tan^{-1} \frac{y - 1}{x - 2} = \ln \left[1 + \frac{\left(y - 1\right)^2}{\left(x - 2\right)^2}\right] + \ln \left(x - 2\right)^2 + \mathrm{constant}$$

However, our assignment didn't come with a standard solution, so I verified my answer with Wolfram Alpha, which gives
$$
2 \tan^{-1}\left(\frac{y(x) + x - 3}{-y(x) + x - 1}\right) = c_1 + \ln\left(\frac{x^2 + y(x)^2 - 2 y(x) - 4 x + 5}{2 \left(x - 2\right)^2}\right) + 2 \ln\left(x - 2\right)$$
which is different from my solution in


*

*the fraction inside function $\tan^{-1}$ is vastly different

*the denominator given by Wolfram Alpha inside the first $\ln$ is twice the denominator I gave

*the $x - 2$ in the last $\ln$ has no absolute value sign around it, but this seems a common problem of Wolfram Alpha solutions, so we can overlook it for the second


May I know whether I'm wrong, or that this is a problem of the Wolfram Alpha solution? (or that the two solutions are actually equivalent, though seemingly very unlikely?)
 A: Solution with free CAS Maxima is
$$\log{\left( {{y}^{2}}-2 y+{{x}^{2}}-4 x+5\right) }+2 \operatorname{atan}\left( \frac{x-2}{y-1}\right) =C$$
or
$$\log{((x-2)^2+(y-1)^2 )}+2 \operatorname{atan}\left( \frac{x-2}{y-1}\right) =C$$
All solutions (yours, Wolfram Alpha and Maxima ) are correct.
Wolfram Alpha use substitution
$$v=\frac{x + y - 3}{x - y - 1}.$$
A: One way to solve
$$
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x+y-3}{x-y-1}
$$
is to let $u=x-2$ and $v=y-1$. Then we get
$$
\frac{\mathrm{d}v}{\mathrm{d}u}=\frac{u+v}{u-v}
$$
which becomes
$$
u\,\mathrm{d}v-v\,\mathrm{d}u=\tfrac12\mathrm{d}\!\left(u^2+v^2\right)
$$
Dividing by $u^2+v^2$ yields
$$
\frac{u\,\mathrm{d}v-v\,\mathrm{d}u}{u^2+v^2}=\frac12\frac{\mathrm{d}\!\left(u^2+v^2\right)}{u^2+v^2}
$$
which is
$$
\mathrm{d}\arctan\left(\tfrac vu\right)=\mathrm{d}\log\sqrt{u^2+v^2}
$$
and thus,
$$
c\,e^{\arctan\left(\frac{y-1}{x-2}\right)}=\sqrt{(x-2)^2+(y-1)^2}
$$
This is the Logarithmic Spiral $r=ce^{\theta}$ centered at $(2,1)$.

A More General Approach
If
$$
\frac{\mathrm{d}y}{\mathrm{d}x}=f\!\left(\frac yx\right)
$$
then
$$
\begin{align}
\frac{\mathrm{d}\frac yx}{\mathrm{d}x}
&=\frac{x\frac{\mathrm{d}y}{\mathrm{d}x}-y}{x^2}\\
&=\frac{f\!\left(\frac yx\right)-\frac yx}{x}
\end{align}
$$
Therefore,
$$
\int\frac{\mathrm{d}\frac yx}{f\!\left(\frac yx\right)-\frac yx}
=\log(x)
$$
Applying this to the question, after $u=x-2$ and $v=y-1$,
$$
\begin{align}
\log(u)
&=\int\frac{\mathrm{d}\frac vu}{\frac{1+\frac vu}{1-\frac vu}-\frac vu}\\
&=\int\frac{\left(1-\frac vu\right)\mathrm{d}\frac vu}{1+\left(\frac vu\right)^2}\\[9pt]
&=\arctan\left(\frac vu\right)-\frac12\log\left(1+\left(\frac vu\right)^2\right)+C
\end{align}
$$
Therefore,
$$
\sqrt{u^2+v^2}=c\,e^{\arctan\left(\frac vu\right)}
$$
A: We bsubstitute $$x=t+2,y=v+1$$
then we have
$$\frac{dv(t)}{dt}=\frac{gt+v(t))}{t-v(t)}$$
then we Substitute:
$$v(t)=tu(t)$$
anhd we get
$$\frac{\frac{du(t)}{dt}(u(t)-1)}{-u(t)^2-1}=\frac{1}{t}$$
A: Attached a plot showing the solutions
$$
S_1\to 2 \tan ^{-1}\left(\frac{x+y-3}{x-y-1}\right)=\log \left(\frac{x^2-4 x+y^2-2 y+5}{2 (x-2)^2}\right)+2 \log (x-2)+C_0\\
S_2\to \tan ^{-1}\left(\frac{y-1}{x-2}\right)-\frac{1}{2} \log \left(\frac{(y-1)^2}{(x-2)^2}+1\right)=\log (x-2)+C_1
$$
with $C_0=0, C_1 = -1.13$
I hope this helps.
In red $S_1$ and in blue $S_2$

