Boundary Value Problems and Solutions

Boundary Value Problems and Solutions

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

Created by

Ethan Morris

FREE Resource

This video introduces boundary value problems (BVPs), explaining their definition and how they differ from initial value problems. It provides two examples of solving BVPs using given general solutions to differential equations. The first example involves exponential functions, while the second uses trigonometric functions. The video demonstrates how to apply initial conditions to find particular solutions and verify them.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the primary objective of the lesson on boundary value problems?

To define and solve boundary value problems

To understand the applications of calculus

To explore the history of differential equations

To learn about initial value problems

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is a boundary value problem different from an initial value problem?

It does not require any conditions

It is solved using algebra

It involves only first-order equations

It has conditions at multiple points

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the first example, what is the general solution provided for the differential equation?

y(x) = c1 * ln(x) + c2 * ln(x^2)

y(x) = c1 * x + c2 * x^2

y(x) = c1 * e^x + c2 * e^x

y(x) = c1 * sin(x) + c2 * cos(x)

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What condition is used to find the relationship between c1 and c2 in the first example?

y'(1) = 1

y(0) = 0

y'(0) = 0

y(1) = 1

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the particular solution found for the first example?

y(x) = (e^x - e^x) / (e - 1 + e)

y(x) = (e^x - e^x) / (e - 1 - e)

y(x) = (e^x + e^x) / (e - 1 + e)

y(x) = (e^x + e^x) / (e - 1 - e)

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the second example, what is the general solution provided for the differential equation?

y(x) = c1 * ln(x) + c2 * ln(x^2)

y(x) = c1 * cos(2x) + c2 * sin(2x)

y(x) = c1 * e^x + c2 * e^x

y(x) = c1 * x + c2 * x^2

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the initial condition used to find c1 in the second example?

y(0) = 1

y'(1) = 0

y'(0) = 1

y(1) = 0

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