# Solve Initial Value Problem and Determining Interval

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1. Oct 6, 2015

### Sylvia

1. The problem statement, all variables and given/known data
Solve the initial value problem and determine at least approximately where the solution is valid

(2x-y) + (2y-x)y' = 0
y(1) = 3

2. Relevant equations

3. 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?

2. Oct 6, 2015

### Staff: Mentor

Actually, the solution is $y = \frac x 2 \pm \frac{\sqrt{28 - 3x^2}}{2}$
IOW, there are two solutions for y in terms of x. Due to the initial condition, y(1) = 3, you choose the pos. square root above.
In order for y to be real, the quantity under the radical has to be nonnegative, so you need to solve the quadratic inequality $28 - 3x^2 \ge 0$. Solving this results in $|x| \le \sqrt{28/3}$, which is slightly different from what you report is the book's answer.