Solving Initial Value Problem: 1st Order Dif Eq Help with 2 Variables

[footballxpaul]

Homework Statement

(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 doesn't 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?

 

[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.

 

[footballxpaul]

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

 

[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.

 

[Дьявол]

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

 

[HallsofIvy]

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

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.

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

 

[footballxpaul]

solved, thanks guys, can't believe I couldn't see those