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?)
