Solve Initial Value Problem and Determining Interval

In summary, the conversation discusses solving an initial value problem and determining the validity of the solution. The solution involves finding the roots of a quadratic equation and using the initial condition to choose the correct solution. The interval of validity is found by setting the quantity under the radical to be nonnegative and solving for x, which results in ##|x| \le \sqrt{28/3}##. However, the book's answer is slightly different, stating that the interval of validity is ##|x| < \sqrt{28/3}##. The general method for finding the interval of validity involves solving for the values of x that make the solution real and valid.
  • #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?
 
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  • #2
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?
 

Related to Solve Initial Value Problem and Determining Interval

1. What is an initial value problem?

An initial value problem is a type of differential equation that involves finding a function or set of functions that satisfy a given set of initial conditions. These initial conditions typically include a starting point or value for the function(s) in question, and may also include information about the rate of change or slope at that point.

2. How do you solve an initial value problem?

To solve an initial value problem, you must first find the general solution to the differential equation. This involves using techniques such as separation of variables, substitution, or integration. Once the general solution is found, you can then use the given initial conditions to find the specific solution that satisfies both the equation and the initial values.

3. What is the interval of a solution to an initial value problem?

The interval of a solution to an initial value problem refers to the range of values for the independent variable (usually denoted as x) over which the solution is defined. This interval is typically determined by the given initial conditions and any restrictions or limitations on the function(s) in the problem.

4. How do you determine the interval of a solution to an initial value problem?

To determine the interval of a solution to an initial value problem, you must first solve the differential equation to find the general solution. Then, you can use the initial conditions to find the specific solution. The interval of the solution will be determined by any restrictions on the function(s) in the problem, as well as the values of the independent variable that satisfy the initial conditions.

5. Why is it important to determine the interval of a solution to an initial value problem?

Determining the interval of a solution to an initial value problem is important because it ensures that the solution is valid and meaningful within a specific range of values. Without this information, the solution may not accurately represent the behavior of the function(s) in question, and may lead to incorrect conclusions or predictions. Additionally, knowing the interval can help identify any potential issues or limitations with the solution, such as points of discontinuity or undefined values.

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