Solve Initial Value Problem and Determining Interval

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SUMMARY

The discussion centers on solving the initial value problem given by the equation (2x-y) + (2y-x)y' = 0 with the initial condition y(1) = 3. The correct solution is derived as y = x/2 ± sqrt(28 - 3x^2)/2, leading to two potential solutions for y in terms of x. The interval of validity is determined by ensuring the expression under the square root remains non-negative, resulting in the condition |x| < sqrt(28/3). This condition is essential for maintaining real values for y.

PREREQUISITES
  • Understanding of differential equations, specifically first-order equations.
  • Familiarity with the quadratic formula and its application.
  • Knowledge of inequalities and their implications on solution validity.
  • Basic calculus concepts, including limits and continuity.
NEXT STEPS
  • Study the derivation of intervals of validity for solutions to differential equations.
  • Learn about the implications of initial conditions on solution behavior.
  • Explore quadratic inequalities and their graphical interpretations.
  • Investigate other methods for solving first-order differential equations, such as separation of variables.
USEFUL FOR

Students studying differential equations, mathematics educators, and anyone interested in understanding the behavior of solutions to initial value problems.

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?
 

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