The video tutorial explains how to solve a linear second order homogeneous differential equation with constant coefficients. It covers setting up the characteristic equation, solving it using the quadratic formula, and determining the general solution. The tutorial then applies initial conditions to find the particular solution and verifies it by calculating the derivative. The final solution is graphed to confirm it meets the initial conditions.

What type of differential equation is discussed in the video?

• Linear first order

• Linear second order homogeneous

• Non-linear second order

• Non-linear first order

Tags

CCSS.HSA-REI.B.4B

What is the form of the characteristic equation for the given differential equation?

• R^2 + R + 2 = 0

• R^2 - 2R + 1 = 0

• R^2 + 2R + 1 = 0

• R^2 - R + 2 = 0

Tags

CCSS.HSA-REI.B.4B

Why is the quadratic formula used to solve the characteristic equation?

• Because the equation is not factorable

• Because the equation is non-linear

• Because the equation is linear

• Because the equation has real roots

Tags

CCSS.HSA-REI.B.4B

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

• C1 * e^(beta*x) + C2 * e^(alpha*x)

• C1 * e^(alpha*x) * cos(beta*x) + C2 * e^(alpha*x) * sin(beta*x)

• C1 * e^(alpha*x) + C2 * e^(beta*x)

• C1 * e^(alpha*x) * sin(beta*x) + C2 * e^(beta*x) * cos(alpha*x)

Tags

CCSS.HSF.TF.B.7

What is the value of alpha in the general solution?

• 2

• 1

• -2

• -1

How is C1 determined using the initial conditions?

• By setting y'(0) = 1

• By setting y(0) = 1

• By setting y(0) = 0

• By setting y'(0) = 0

Tags

CCSS.HSF.BF.A.2

What is the value of C2 after applying the initial conditions?

• 3 / sqrt(7)

• 1 / sqrt(7)

• 4 / sqrt(7)

• 2 / sqrt(7)

What is the final form of the solution y(x)?

• y(x) = 4 / sqrt(7) * e^(x/2) * sin(sqrt(7)/2 * x)

• y(x) = 2 / sqrt(7) * e^(x/2) * cos(sqrt(7)/2 * x)

• y(x) = 2 / sqrt(7) * e^(x/2) * sin(sqrt(7)/2 * x)

• y(x) = 4 / sqrt(7) * e^(x/2) * cos(sqrt(7)/2 * x)

Tags

CCSS.HSF-IF.C.7E

What does the graph of the solution indicate about the initial conditions?

• The slope at x=0 is zero

• The slope at x=0 is negative

• The function passes through the origin

• The function does not pass through the origin

Tags

CCSS.8.EE.B.5

What is the slope of the tangent line at x=0 according to the graph?

• Positive 2

• Zero

• Positive 1

• Negative 2

Tags

CCSS.8.EE.B.5