How can I solve this differental equation? \(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{l}{i}{g}{n}\right\rbrace}{\left({\frac{{{1}}}{{{x}}}}-{y}^{{2}}{\frac{{{1}}}{{{\left({x}-{y}\right)}^{{2}}}}}\right)}{\left.{d}{x}\right.}&+{\left({x}^{{2}}{\frac{{{1}}}{{{\left({x}-{y}\right)}^{{2}}}}}-{\frac{{{1}}}{{{y}}}}\right)}{\left.{d}{y}\right.}={0}\quad?\backslash{e}{n}{d}{\left\lbrace{a}{l}{i}{g}{n}\right\rbrace}\) I

amonitas3zeb

amonitas3zeb

Answered question

2022-03-27

How can I solve this differental equation?
(1xy21(xy)2)dx+(x21(xy)21y)dy=0 ?
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.

Answer & Explanation

Alannah Campos

Alannah Campos

Beginner2022-03-28Added 10 answers

Step 1
The equation can be rewritten as
dxx-y2(x-y)2dx+x2(x-y)2dy-dyy=0.
Step 2
Here, I would suggest the substitution xw=y. Hence

dxxw2(1w)2dx+1(1w)2d(xw)d(xw)xw

=dxxw2(1w)2dx+x(1w)2dw+w(1w)2dxdwwdxx

=w2(1w)2dx+w(1w)2dx+x(1w)2dwdww

=w(1w)(1w)2dx+xd(11w)d[ln(w)]

=w1wdx+xd(11w)d[ln(w)]

=dx+dx1w+xd(11w)d[ln(w)]

=d(x1w)dxd[ln(w)]

=d[x1wxln(w)]

=d[x(xxw)1wln(w)]

=d[xw1wln(w)]

=d[x2wxxwln(w)]

=d[xyxyln(y)+ln(x)]

=0

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