Differential Equation ?

  • Thread starter thorjj
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  • #1
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Homework Statement



Solve (x^2 + y^2 - y)dx + x dy = 0 when x > 0, y > 0.

Homework Equations



(x^2 + y^2 - y)dx + x dy = 0

The Attempt at a Solution



dy/dx = - (x^2 + y^2 - y) / x
That seems like a tricky one. Can I make it a second-order diff. eq.?
Well, with the topic we have had about vectorfields I can backwards engineer it into
F(x,y) = (x^2 + y^2 - y)i + (-x)j
But that dosn't make much sense to me.
 

Answers and Replies

  • #2
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(Hmm, I miss an edit-button...)

I shorted it down to x dy/dx = -x^2 - y^2 + y or dy/dx = -x -y^2/x + y/x
 
  • #3
Defennder
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You don't have to complicate this one unnecessarily. The substitution y=vx works fine here.
 
  • #4
Shooting Star
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[tex](x^2+y^2)dx + (xdy-ydx) = 0 =>
dx + \frac{xdy-ydx}{x^2+y^2} = 0.[/tex]

Whenever you get (xdy-ydx), see if you can make it into the form

[tex]\frac{xdy-ydx}{x^2}[/tex], since that is equal to [tex]d(\frac{y}{x}).[/tex]

See if you can spot a function of (y/x) with the d(y/x) here.
 

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