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1st order homogenous ODE: (x+y)dy/dx=(x-y)

  • Thread starter 2h2o
  • Start date
  • #1
53
0

Homework Statement


Find a general solution


Homework Equations


[tex] (x+y)\frac{dy}{dx} = x-y [/tex]


The Attempt at a Solution


[tex]

\frac{dy}{dx} = \frac{x-y}{x+y}

[/tex]

let v=y/x
y=xv

[tex]

\frac{dy}{dx} = v+x\frac{dv}{dx}

[/tex]

now,

[tex]

v+x\frac{dv}{dx} = \frac{x-xv}{x+xv} [/tex]

[tex]= \frac{1-v}{1+v} [/tex]

[tex]= \frac{1}{1+v} - \frac{v}{1+v} [/tex]

[tex]=\frac{1}{1+\frac{y}{x}}-\frac{\frac{y}{x}}{1+\frac{y}{x}}[/tex]

Which takes me back to where I started if I clear the denominators, so I'm spinning the wheels. This looks very familiar to me, but I don't recall what it is or what to do with it. Separate the variables, then integrate? Could do it, wrt v, by parts, but isn't there a more efficient way?
 
Last edited:

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
14
Well, first, v/(1+v) = ((v+1)-1)/(1+v) = 1 - 1/(v+1). So that will help.
 
  • #3
53
0
Ok, that gets me a little further, but I still run into trouble:

[tex]
= \frac{1}{1+v} - \frac{v}{1+v}
[/tex]

[tex]
= \frac{1}{1+v} - \frac{(1+v)-1}{1+v}
[/tex]

[tex]
= \frac{1}{1+v} - 1 + \frac{1}{1+v}
[/tex]

[tex]
= \frac{2}{1+v} - 1
[/tex]

[tex]
= \frac{2-(1+v}{1+v}
[/tex]

[tex]
= \frac{2}{1+v} - 1
[/tex]

Which will lead me again, back to my starting point. When do I actually start back-substituting to get something useful?
 
  • #4
53
0
Problem solved: way back in one of the first steps, I found it to be much easier if I collect all the v's on the RHS (but leaving x*dv/dx on the LHS) and onwards from there. Thanks Char, for trying to help anyway :)
 

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