Differential Equations with Repeated Roots

Differential Equations with Repeated Roots

Assessment

Interactive Video

Mathematics

10th - 12th Grade

Hard

Created by

Liam Anderson

FREE Resource

The video tutorial explains how to solve a second-order linear homogeneous differential equation with repeated roots. It begins by introducing the problem and deriving the characteristic equation. The instructor uses the quadratic formula to find the repeated root and discusses the need for a more general solution due to the initial conditions. The video then demonstrates how to apply these initial conditions to determine specific constants, ultimately deriving the particular solution. The tutorial concludes with a brief introduction to solving non-homogeneous differential equations.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the initial condition given for y(0) in the problem?

y(0) = 3

y(0) = 4

y(0) = 1

y(0) = 2

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the characteristic equation derived from the given differential equation?

r^2 + r + 0.25

r^2 - r + 0.25

r^2 - r - 0.25

r^2 + r - 0.25

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is a more general solution needed for repeated roots?

Because the initial solution is too complex

Because the initial solution is not general enough for two initial conditions

Because the initial solution is incorrect

Because the initial solution is too simple

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the form of the general solution for repeated roots?

y = c1 * e^(1/2 * x) + c2 * x * e^(1/2 * x)

y = c1 * e^(x)

y = c1 * e^(1/2 * x)

y = c1 * x * e^(1/2 * x)

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the derivative of the general solution used to find c1 and c2?

y' = c1 * e^(1/2 * x)

y' = 1/2 * c1 * e^(1/2 * x) + c2 * e^(1/2 * x) + 1/2 * c2 * x * e^(1/2 * x)

y' = c2 * e^(x)

y' = c1 * x * e^(1/2 * x)

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the value of c1 found using the initial conditions?

c1 = 3

c1 = 2

c1 = 4

c1 = 1

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the value of c2 found using the initial conditions?

c2 = -1/3

c2 = 2/3

c2 = 1/3

c2 = -2/3

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