Solving a Tricky Second-Order Differential Equation

[thorjj]

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.

 

[thorjj]

(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

 

[Defennder]

You don't have to complicate this one unnecessarily. The substitution y=vx works fine here.

 

[Shooting Star]

$$(x^2+y^2)dx + (xdy-ydx) = 0 =><br /> dx + \frac{xdy-ydx}{x^2+y^2} = 0.$$

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

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

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