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.

(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)

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

step-by-step explanation:

6.48/4 4 goes into 6.48 1.62 times