# 1st order dif eq help

1. Jul 15, 2009

### footballxpaul

1. The problem statement, all variables and given/known data

(2x-y)dx+(2y-x)dy=0 y(1)=3

solve the given initial value problem and determine at least approx where the solution is valid?

this should be simple right? I got 2y+2x=C and then 2y+2x=8. It doesnt match the answer in the back of the book at all, and I do see how they could have gotten the answer in the back. Is my answer right? The book says y=[x+sqrt(28-3x^2)]/2, abs(x)<sqrt(28/3)? or is the book right, and if so how do you get to their answer?

2. Jul 15, 2009

### snipez90

The book is right, this is an exact differential equation. You can tell this since the partial derivative of the term multiplied by dx, namely (2x-y), with respect to y is equal to the partial derivative of the term multiplied by dy, or (2y - x), with respect to x (since both partials are -1).

Check out http://www.sosmath.com/diffeq/first/exact/exact.html. Follow the method step by step and that will lead you straight to the solution.

3. Jul 15, 2009

### footballxpaul

thanks I got to x^2-yx+y^2=7 now. I have no clue how to seperate this right now, is there a trick, did I do something wrong.

4. Jul 15, 2009

### snipez90

No you basically got this. To complete the problem, subtract 7 from both sides and view the equation as a quadratic in y (treating other quantities as constants) and apply the quadratic formula.

5. Jul 15, 2009

### Дьявол

And how did you solve this
(2x-y)dx+(2y-x)dy=0
?

6. Jul 16, 2009

### HallsofIvy

Staff Emeritus
But, generally speaking, it is not necessary to "solve" the equation for y. $x^2- xy+ y^2=7$ is a perfectly good solution.

7. Jul 16, 2009

### footballxpaul

solved, thanks guys, cant believe I couldnt see those

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