amonitas3zeb

2022-03-27

How can I solve this differental equation?

$\begin{array}{rl}(\frac{1}{x}-{y}^{2}\frac{1}{(x-y{)}^{2}})dx& +({x}^{2}\frac{1}{(x-y{)}^{2}}-\frac{1}{y})dy=0\phantom{\rule{1em}{0ex}}?\end{array}$

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

Beginner2022-03-28Added 10 answers

Step 1

The equation can be rewritten as

$\frac{{\displaystyle \mathrm{d}x}}{{\displaystyle x}}-\frac{{\displaystyle {y}^{2}}}{{\displaystyle (x-y{)}^{2}}}\mathrm{d}x+\frac{{\displaystyle {x}^{2}}}{{\displaystyle (x-y{)}^{2}}}\mathrm{d}y-\frac{{\displaystyle \mathrm{d}y}}{{\displaystyle y}}=0.$

Step 2

Here, I would suggest the substitution $xw=y$. Hence

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

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

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

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

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

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

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

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

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

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

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

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

$=0$

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