Differential Equations Concepts Review

Differential Equations Concepts Review

Assessment

Interactive Video

Mathematics

10th - 12th Grade

Hard

CCSS
HSA.REI.A.2

Standards-aligned

Created by

Mia Campbell

FREE Resource

Standards-aligned

CCSS.HSA.REI.A.2
The video tutorial explains solving an initial value problem using substitution. It starts by identifying the need for a general substitution due to the presence of a square root in the differential equation. By letting V equal x squared plus y squared, the equation is transformed into a separable differential equation. The tutorial then demonstrates integrating both sides to solve for V, and subsequently rewriting the equation in terms of X and Y. The particular solution is found using initial conditions, and the solution is verified graphically. The tutorial concludes with a simplified form of the solution.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main challenge in solving the given differential equation?

It has constant coefficients.

It is a Bernoulli equation.

It is a homogeneous equation.

The presence of a square root term.

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What substitution is used to simplify the differential equation?

V = y^2 - x^2

V = x^2 - y^2

V = x + y

V = x^2 + y^2

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

After substitution, what type of differential equation is obtained?

Linear

Separable

Exact

Non-linear

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the result of integrating the equation with respect to V?

2V^(1/2)

V^(1/2)

V^2

V^(1/2)/2

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the constant of integration represented in the solution?

C

C sub 3

C sub 2

C sub 1

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What initial condition is used to find the particular solution?

y(0) = 6

y(0) = 2

y(0) = 0

y(0) = 4

Tags

CCSS.HSA.REI.A.2

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the simplified form of the particular solution?

y = sqrt(x + 2)

y = sqrt(8x + 16)

y = sqrt(4x + 8)

y = sqrt(2x + 4)

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