Differential Equations and Initial Value Problems

Differential Equations and Initial Value Problems

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

Created by

Amelia Wright

FREE Resource

This video tutorial explores the existence and uniqueness of solutions to initial value problems using Picard's theorem. It explains the conditions under which a solution exists and whether it is unique. Through three examples, the video demonstrates scenarios where solutions do not exist, are not unique, and are unique, using slope fields and mathematical analysis to illustrate each case.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the two fundamental questions addressed in the study of initial value problems?

Does a solution exist and is it unique?

What are the roots and factors of the equation?

What is the derivative and integral of the function?

How to find the maximum and minimum values?

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

According to Picard's theorem, what must be true for a solution to exist and be unique?

The function must be differentiable and integrable.

The function and its partial derivative with respect to y must be continuous.

The function must be linear and homogeneous.

The function must have a constant rate of change.

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the example y' = 1/x, why does a solution not exist at the initial condition y(0) = 0?

The function is linear at x = 0.

The function is not differentiable at x = 0.

The function is not continuous at x = 0.

The function has a maximum at x = 0.

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the general solution for the differential equation y' = 1/x?

y = e^x + C

y = x + C

y = ln|x| + C

y = x² + C

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the example y' = x√(y-2), why is the solution not unique?

The function has multiple roots.

The function is not defined for x = 1.

The partial derivative with respect to y is not continuous at y = 2.

The function is not continuous at y = 2.

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the two solutions for the initial value problem y' = x√(y-2) with y(1) = 2?

y = x² and y = 2

y = x³ and y = 3

y = 2 and y = x²

y = x and y = 1

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

For the differential equation y' = xy², what indicates that the solution is unique?

The function and its partial derivative with respect to y are continuous.

The function is linear.

The function is quadratic.

The function has a constant solution.

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