- #1
Sylvia
- 30
- 1
Homework Statement
Solve the initial value problem and determine at least approximately where the solution is valid
(2x-y) + (2y-x)y' = 0
y(1) = 3
Homework Equations
The Attempt at a Solution
I know how to solve it, and I got the correct answer, which was:
7 = x^2 - yx + y^2
and then applying the quadratic formula
x+sqrt(28 - 3x^2) / (2)
However, I have trouble determining where the solution is valid. I understand that to find it you set 28 - 3x^2 = 0, and then solve for x (which comes out to x = +/- sqrt(28/3). However, the answers in the back of the book say that the interval of validity is IxI < sqrt(28/3). Can anyone explain how this is derived? Why is it x < sqrt(28/3)?Also, could someone please explain to me how to find the interval of validity in general?