Solve Initial Value Problem and Determining Interval

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The initial value problem given is (2x-y) + (2y-x)y' = 0 with the condition y(1) = 3. The correct solution derived is y = (x/2) ± (sqrt(28 - 3x^2)/2), with the positive root chosen due to the initial condition. To determine the interval of validity, the expression under the square root must be nonnegative, leading to the inequality 28 - 3x^2 ≥ 0, which simplifies to |x| ≤ sqrt(28/3). This indicates that the solution is valid for x in the interval (-sqrt(28/3), sqrt(28/3)). Understanding the interval of validity involves ensuring the conditions for real solutions are met within the derived constraints.
Sylvia
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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?
 
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Sylvia said:

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)
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.
Sylvia said:
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).
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.
Sylvia said:
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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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