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Match the third order linear equations with their fundamental solution sets. 1. y′′′−6y′′+y′−6y=0 2. y′′′−8y′′+15y′=0 3. y′′′+y′=0 4. y′′′−y

Home/Math/Match the third order linear equations with their fundamental solution sets. 1. y′′′−6y′′+y′−6y=0 2. y′′′−8y′′+15y′=0 3. y′′′+y′=0 4. y′′′−y

Match the third order linear equations with their fundamental solution sets. 1. y′′′−6y′′+y′−6y=0 2. y′′′−8y′′+15y′=0 3. y′′′+y′=0 4. y′′′−y

Question

Match the third order linear equations with their fundamental solution sets. 1. y′′′−6y′′+y′−6y=0 2. y′′′−8y′′+15y′=0 3. y′′′+y′=0 4. y′′′−y′′−y′+y=0 5. ty′′′−y′′=0 6. y′′′+3y′′+3y′+y=0

The fundamental solution sets to the third order linear equations are not given. I would however, find the solutions, so that you can match them to their corresponding options.

Answer:

(1) y”’ – 6y” + y’ – 6y = 0

Corresponds to:

y = C1e^(6t) + C2cost + C3sint

(2) y”’ – 8y” + 15y’ = 0

Corresponds to:

y = C1 + C2e^(3t) + C3e^(5t)

(3) y”’ + y’ = 0

Corresponds to:

y = C1 + C2cost + C3sint

(4) y”’ – y” – y’ + y = 0

Corresponds to:

y = C1e^(-t) + (C2 + C3t) e^t

(5) ty”’ – y” = 0

Corresponds to:

y = C1 + C2t + C3t³

(6) y”’ + 3y” + 3y’ + y = 0

Corresponds to:

y = (C1 + C2t + C3t²)e^(-t)

Step-by-step explanation:

Given the following differential equations

(1) y”’ – 6y” + y’ – 6y = 0

(2) y”’ – 8y” + 15y’ = 0

(3) y”’ + y’ = 0

(4) y”’ – y” – y’ + y = 0

(5) ty”’ – y” = 0

(6) y”’ + 3y” + 3y’ + y = 0

SOLUTIONS

(1) y”’ – 6y” + y’ – 6y = 0

We write the characteristic equation and solve.

The characteristic equation is

m³ – 6m² + m – 6 = 0

m²(m – 6) + (m – 6) = 0

(m² + 1)(m – 6) = 0

m² + 1 = 0

=> m² = -1

=> m = ±√(-1) = ±i

Or m – 6 = 0

=> m = 6

The solutions are therefore

m = ±i, 6

Therefore, the solution to the differential equation is

## Answers ( )

The fundamental solution sets to the third order linear equations are not given. I would however, find the solutions, so that you can match them to their corresponding options.

Answer:

(1) y”’ – 6y” + y’ – 6y = 0

Corresponds to:

y = C1e^(6t) + C2cost + C3sint

(2) y”’ – 8y” + 15y’ = 0

Corresponds to:

y = C1 + C2e^(3t) + C3e^(5t)

(3) y”’ + y’ = 0

Corresponds to:

y = C1 + C2cost + C3sint

(4) y”’ – y” – y’ + y = 0

Corresponds to:

y = C1e^(-t) + (C2 + C3t) e^t

(5) ty”’ – y” = 0

Corresponds to:

y = C1 + C2t + C3t³

(6) y”’ + 3y” + 3y’ + y = 0

Corresponds to:

y = (C1 + C2t + C3t²)e^(-t)

Step-by-step explanation:

Given the following differential equations

(1) y”’ – 6y” + y’ – 6y = 0

(2) y”’ – 8y” + 15y’ = 0

(3) y”’ + y’ = 0

(4) y”’ – y” – y’ + y = 0

(5) ty”’ – y” = 0

(6) y”’ + 3y” + 3y’ + y = 0

SOLUTIONS

(1) y”’ – 6y” + y’ – 6y = 0

We write the characteristic equation and solve.

The characteristic equation is

m³ – 6m² + m – 6 = 0

m²(m – 6) + (m – 6) = 0

(m² + 1)(m – 6) = 0

m² + 1 = 0

=> m² = -1

=> m = ±√(-1) = ±i

Or m – 6 = 0

=> m = 6

The solutions are therefore

m = ±i, 6

Therefore, the solution to the differential equation is

y = C1e^(6t) + C2cost + C3sint

(2) y”’ – 8y” + 15y’ = 0

The characteristic equation is

m³ – 8m² + 15m = 0

m³ – 5m² – 3m² + 15m = 0

m²(m – 5) – 3m(m – 5) = 0

(m² – 3m)(m – 5) = 0

m(m – 3)(m – 5) = 0

m = 0

Or

m – 3 = 0

=> m = 3

Or

m – 5 = 0

=> m = 5

Therefore,

y = C1 + C2e^(3t) + C3e^(5t)

(3) y”’ + y’ = 0

The characteristic equation is

m³ + m = 0

m(m² + 1) = 0

m = 0

Or

m² + 1 = 0

=> m² = -1

=> m = ±√(-1) = ±i

Therefore,

y = C1 + C2cost + C3sint

(4) y”’ – y” – y’ + y = 0

The characteristic equation is

m³ – m² – m + 1 = 0

m²(m – 1) – (m – 1) = 0

(m² – 1)(m – 1) = 0

(m – 1)(m + 1)(m – 1) = 0

Then

m – 1 = 0 twice

=> m = 1 twice

Or

m + 1 = 0

=> m = -1

Therefore,

y = C1e^(-t) + (C2 + C3t) e^t

(5) ty”’ – y” = 0

Multiplying by t², this can be written as

t³y”’ – t²y” = 0

Put t = e^z, so that z = lnt, and Dy = tdy/dt

Then

dy/dt = (1/t)dy/dz

ty’ = dy/dz

Again,

d²y/dt² = -(1/t²)dy/dz + (1/t)(d²y/dz²)(dz/dt)

=> t²y” = D(D – 1)y

Similarly,

t³y”’ = D(D – 1)(D – 2)y

Using these, we have

[D(D – 1)(D – 2) – D(D – 1)]y = 0

D(D – 1)(D – 3)y = 0

The characteristic equation is

m(m – 1)(m – 3) = 0

m = 0

Or

m – 1 = 0

=> m = 1

Or

m – 3 = 0

=> m = 3

y = C1 + C2e^z + C3(e^z)³

But e^z = t

So

y = C1 + C2t + C3t³

(6) y”’ + 3y” + 3y’ + y = 0

The characteristic equation is

m³ + 3m² + 3m + 1 = 0

(m + 1)³ = 0

m = -1 three times

Therefore,

y = (C1 + C2t + C3t²)e^(-t)