 amonitas3zeb

2022-03-27

How can I solve this differental equation?

I have this as a part of my homework, and there were like 20 other differential equations that I easily solved, but this one stood out.
This is not linear, not separable. I don't know how to approach this problem. If you could tell me to what class this kind of equation belongs, and some methods to solve them, I would be very very glad. Alannah Campos

Step 1
The equation can be rewritten as
$\frac{\mathrm{d}x}{x}-\frac{{y}^{2}}{\left(x-y{\right)}^{2}}\mathrm{d}x+\frac{{x}^{2}}{\left(x-y{\right)}^{2}}\mathrm{d}y-\frac{\mathrm{d}y}{y}=0.$
Step 2
Here, I would suggest the substitution $xw=y$. Hence

$\frac{\mathrm{d}x}{x}-\frac{{w}^{2}}{\left(1-w{\right)}^{2}}\mathrm{d}x+\frac{1}{\left(1-w{\right)}^{2}}\mathrm{d}\left(xw\right)-\frac{\mathrm{d}\left(xw\right)}{xw}$

$=\frac{\mathrm{d}x}{x}-\frac{{w}^{2}}{\left(1-w{\right)}^{2}}\mathrm{d}x+\frac{x}{\left(1-w{\right)}^{2}}\mathrm{d}w+\frac{w}{\left(1-w{\right)}^{2}}\mathrm{d}x-\frac{\mathrm{d}w}{w}-\frac{\mathrm{d}x}{x}$

$=-\frac{{w}^{2}}{\left(1-w{\right)}^{2}}\mathrm{d}x+\frac{w}{\left(1-w{\right)}^{2}}\mathrm{d}x+\frac{x}{\left(1-w{\right)}^{2}}\mathrm{d}w-\frac{\mathrm{d}w}{w}$

$=\frac{w\left(1-w\right)}{\left(1-w{\right)}^{2}}\mathrm{d}x+x\mathrm{d}\left(\frac{1}{1-w}\right)-\mathrm{d}\left[\mathrm{ln}\left(w\right)\right]$

$=\frac{w}{1-w}\mathrm{d}x+x\mathrm{d}\left(\frac{1}{1-w}\right)-\mathrm{d}\left[\mathrm{ln}\left(w\right)\right]$

$=-\mathrm{d}x+\frac{\mathrm{d}x}{1-w}+x\mathrm{d}\left(\frac{1}{1-w}\right)-\mathrm{d}\left[\mathrm{ln}\left(w\right)\right]$

$=\mathrm{d}\left(\frac{x}{1-w}\right)-\mathrm{d}x-\mathrm{d}\left[\mathrm{ln}\left(w\right)\right]$

$=\mathrm{d}\left[\frac{x}{1-w}-x-\mathrm{ln}\left(w\right)\right]$

$=\mathrm{d}\left[\frac{x-\left(x-xw\right)}{1-w}-\mathrm{ln}\left(w\right)\right]$

$=\mathrm{d}\left[\frac{xw}{1-w}-\mathrm{ln}\left(w\right)\right]$

$=\mathrm{d}\left[\frac{{x}^{2}w}{x-xw}-\mathrm{ln}\left(w\right)\right]$

$=\mathrm{d}\left[\frac{xy}{x-y}-\mathrm{ln}\left(y\right)+\mathrm{ln}\left(x\right)\right]$

$=0$

Do you have a similar question?