Substitution in Differential Equations

Substitution in Differential Equations

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

Created by

Lucas Foster

FREE Resource

This video tutorial covers solving differential equations using substitution. It begins with an introduction to the method, followed by a change of variables to simplify the equation. The tutorial then demonstrates substitution and separation of variables for integration, using partial fraction decomposition to solve integrals. Finally, it shows how to solve for the new variable and convert back to the original variables, deriving the final solution and concluding the lesson.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the primary goal of using substitution in solving differential equations?

To find an exact solution

To simplify the equation

To make the equation more complex

To eliminate variables

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the substitution process, what is the new variable V defined as?

The derivative of Y

The product of X and Y

The sum of X and Y

The expression x minus y plus one

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What type of differential equation is obtained after substitution?

Linear

Homogeneous

Non-separable

Separable

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What mathematical technique is used to integrate the simplified equation?

Partial fraction decomposition

Integration by parts

Numerical integration

Trigonometric substitution

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the purpose of performing partial fraction decomposition in this context?

To eliminate constants

To differentiate the equation

To find the roots of the equation

To simplify the integration process

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the constant D introduced in the solution?

As a substitution for e to the power of 2c

As a factor of X

As a derivative of V

As a result of integration

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the two specific solutions obtained for Y in terms of X?

Y = X and Y = X + 1

Y = X - 2 and Y = X + 1

Y = X and Y = X + 2

Y = X - 1 and Y = X + 2

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