Differential Equations Concepts Review

Differential Equations Concepts Review

Assessment

Interactive Video

Mathematics

10th - 12th Grade

Hard

Created by

Olivia Brooks

FREE Resource

The video tutorial explains how to solve an initial value problem involving a linear second order homogeneous differential equation with constant coefficients. It covers the characteristic equation, different forms of general solutions based on the roots, and provides a step-by-step solution to an example problem. The tutorial also discusses the behavior of solutions, particularly focusing on oscillation and amplitude changes based on the real part of complex roots.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What type of differential equation is being solved in this tutorial?

Linear third-order with constant coefficients

Linear first-order with variable coefficients

Non-linear second-order with constant coefficients

Linear second-order with constant coefficients

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the form of the characteristic equation used to solve the differential equation?

a r^2 + br + c = 0

a r + b = 0

a r^3 + br^2 + c = 0

a r^2 + b = 0

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the roots of the characteristic equation when the coefficients are a=1, b=0, and c=25?

0 and 25

1 and -1

5i and -5i

5 and -5

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the general solution form when the characteristic equation has complex roots?

y(t) = c1 e^(alpha t) + c2 e^(-alpha t)

y(t) = c1 e^(beta t) + c2 e^(-beta t)

y(t) = c1 cos(beta t) + c2 sin(beta t)

y(t) = c1 e^(alpha t) cos(beta t) + c2 e^(alpha t) sin(beta t)

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the value of c2 when y(pi/10) = -4?

2

4

-4

0

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the value of c1 when y'(pi/10) = -20?

-4

0

4

2

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the particular solution for the given initial conditions?

y(t) = -4 cos(5t) - 4 sin(5t)

y(t) = -4 cos(5t) + 4 sin(5t)

y(t) = 4 cos(5t) - 4 sin(5t)

y(t) = 4 cos(5t) + 4 sin(5t)

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