Answer:
![\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}](https://tex.z-dn.net/?f=%5Cfrac%7Bu%5E2%2B3%7D%7B-u%5E3%2Bu%5E2-2u%7Du%27%3D%5Cfrac%7B1%7D%7Bx%7D)
Step-by-step explanation:
First step: I'm going to solve our substitution for y:
![u=\frac{y}{x}](https://tex.z-dn.net/?f=u%3D%5Cfrac%7By%7D%7Bx%7D)
Multiply both sides by x:
![ux=y](https://tex.z-dn.net/?f=ux%3Dy)
Second step: Differentiate the substitution:
![u'x+u=y'](https://tex.z-dn.net/?f=u%27x%2Bu%3Dy%27)
Third step: Plug in first and second step into the given equation dy/dx=f(x,y):
![u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}](https://tex.z-dn.net/?f=u%27x%2Bu%3D%5Cfrac%7Bx%28ux%29%2B%28ux%29%5E2%7D%7B3x%5E2%2B%28ux%29%5E2%7D)
![u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}](https://tex.z-dn.net/?f=u%27x%2Bu%3D%5Cfrac%7Bux%5E2%2Bu%5E2x%5E2%7D%7B3x%5E2%2Bu%5E2x%5E2%7D)
We are going to simplify what we can.
Every term in the fraction on the right hand side of equation contains a factor of
so I'm going to divide top and bottom by
:
![u'x+u=\frac{u+u^2}{3+u^2}](https://tex.z-dn.net/?f=u%27x%2Bu%3D%5Cfrac%7Bu%2Bu%5E2%7D%7B3%2Bu%5E2%7D)
Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:
Subtract u on both sides:
![u'x=\frac{u+u^2}{3+u^2}-u](https://tex.z-dn.net/?f=u%27x%3D%5Cfrac%7Bu%2Bu%5E2%7D%7B3%2Bu%5E2%7D-u)
Find a common denominator: Multiply second term on right hand side by
:
![u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}](https://tex.z-dn.net/?f=u%27x%3D%5Cfrac%7Bu%2Bu%5E2%7D%7B3%2Bu%5E2%7D-%5Cfrac%7Bu%283%2Bu%5E2%29%7D%7B3%2Bu%5E2%7D)
Combine fractions while also distributing u to terms in ( ):
![u'x=\frac{u+u^2-3u-u^3}{3+u^2}](https://tex.z-dn.net/?f=u%27x%3D%5Cfrac%7Bu%2Bu%5E2-3u-u%5E3%7D%7B3%2Bu%5E2%7D)
![u'x=\frac{-u^3+u^2-2u}{3+u^2}](https://tex.z-dn.net/?f=u%27x%3D%5Cfrac%7B-u%5E3%2Bu%5E2-2u%7D%7B3%2Bu%5E2%7D)
Third step: I'm going to separate the variables:
Multiply both sides by the reciprocal of the right hand side fraction.
![u' \frac{3+u^2}{-u^3+u^2-2u}x=1](https://tex.z-dn.net/?f=u%27%20%5Cfrac%7B3%2Bu%5E2%7D%7B-u%5E3%2Bu%5E2-2u%7Dx%3D1)
Divide both sides by x:
![\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2Bu%5E2%7D%7B-u%5E3%2Bu%5E2-2u%7Du%27%3D%5Cfrac%7B1%7D%7Bx%7D)
Reorder the top a little of left hand side using the commutative property for addition:
![\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}](https://tex.z-dn.net/?f=%5Cfrac%7Bu%5E2%2B3%7D%7B-u%5E3%2Bu%5E2-2u%7Du%27%3D%5Cfrac%7B1%7D%7Bx%7D)
The expression on left hand side almost matches your expression but not quite so something seems a little off.