Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region. (x2 + y2)y' = y2 1. A unique solution exists in the region y ≥ x.
2. A unique solution exists in the entire xy-plane.
3. A unique solution exists in the region y ≤ x.
4. A unique solution exists in the region consisting of all points in the xy-plane except the origin.
5. A unique solution exists in the region x2 + y2 < 1.
Answers
A unique solution exists in the region consisting of all points in the xy-plane except the origin.
The correct option is 4.
The given differential equation is:
(x² + y²)y' = y²
The equation can be rewritten as:
We need to determine a region of the xy-plane for which the differential equation would have a unique solution whose graph passes through a point (x₀, y₀) in the region.
To determine the region, we can use the existence and uniqueness theorem for first-order differential equations.
According to the theorem, a unique solution exists in a region if the differential equation is continuous and satisfies the Lipschitz condition in that region.
To check if the differential equation satisfies the Lipschitz condition, we can take the partial derivative of the equation with respect to y:
dy/dx = y / (x² + y²)
The partial derivative is continuous and bounded in the entire xy-plane except at the origin (x=0, y=0).
Therefore, the differential equation satisfies the Lipschitz condition in the entire xy-plane except at the origin.
Since the differential equation is continuous in the entire xy-plane, a unique solution exists in any region that does not contain the origin. Therefore, the correct answer is:
A unique solution exists in the region consisting of all points in the xy-plane except the origin.
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Final answer:
The differential equation will have a unique solution in the entire xy-plane except at the origin, as both the function and its partial derivatives are continuous and well-defined everywhere except at that point.
Explanation:
To determine a region of the xy-plane where the differential equation (x2 + y2)y' = y2 has a unique solution passing through a point (x0, y0), we need to consider where the function and its derivative are continuous and well-defined. According to the existence and uniqueness theorem for differential equations, a necessary condition for a unique solution to exist is that the functions of x and y in the equation, as well as their partial derivatives with respect to y, should be continuous in the region around the point (x0, y0).
We note that both the function (x2 + y2)y' and its partial derivative with respect to y, which is 2y, are continuous and well-defined everywhere except at the origin where x = 0 and y = 0. Therefore, a unique solution exists in the region consisting of all points in the xy-plane except the origin.
From the given options, the correct answer is:
4. A unique solution exists in the region consisting of all points in the xy-plane except the origin.
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Related Questions
100 point question.......
What is the value of the expression |5x+12| when x=5
What is the y-intercept of the graph below.
Can someone please help me god bless!
Answers
Answer:
it should be about 6. try measuring the the 3 line with paper or something then put it on the other line
Step-by-step explanation:
Answers
Answer:
Range : [-6, ∞)
Step-by-step explanation:
Domain of any function on a graph is represented by the x-values (input values).
Similarly, Range of function is represented by the y-values or output values of the function on a graph.
Therefore, domain of the given absolute function will be (-∞, ∞) Or set of all real numbers.
Range of the function → [-6, ∞) Or {y | y ≥ -6}
Answers
Those lengths have a common factor of 3. Removing that factor gives you the smaller similar triangle with sides 2, 3, and 4.
2
while Philip says that the slope is 2.
Which reason correctly justifies Tom's answer?
Answers
Answer:
the answer is d
Step-by-step explanation:
i took it on usatestprep.
Answers
Answer:
An employee's score in order to qualify for increase in the salary must be higher than 95.45.
Step-by-step explanation:
Let X represent the performance score of employees.
It is provided that X follows a normal distribution with parameters μ = 82.5 and σ - 9.25.
It is provided that the office will increase salary of its top 8% employees on the basis of a performance score the office created for each employee.
That is, the probability to qualify for increase in the salary is,
P (X > x) = 0.08
⇒ P (X < x) = 0.92
⇒ P (Z < z) = 0.92
The corresponding z-value is,
z = 1.40
Compute the value of x as follows:
Thus, an employee's score in order to qualify for increase in the salary must be higher than 95.45.
Answers
The supplement of <DCE is <ACE, remember, a supplementary pair is a pair of two angles whose sum of degrees sums up to 180 degrees.
The verticle angle to <BCD is <ACF. A verticle angle is formed when two lines intersect, there are four angles formed, the verticle angles are the angles opposite to each other, and verticle angles are congruent.