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.
(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)
