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3 years ago

Find a cubic function that has the roots 5 and 3-2i

Mathematics

1 answer:

Zolol [24]3 years ago

4 0

Answer:

$P(x)=x^3-11x^2+43x-65$

Step-by-step explanation:

If the complex number $3-2i$ is a root of a cubic function, then the complex number $3+2i$ is a root too. Thus, the cubic function has three known roots $5,\ 3-2i,\ 3+2i$ and can be written as

$P(x)=(x-5)(x-(3-2i))(x-(3+2i)),\\ \\P(x)=(x-5)(x^2-x(3-2i+3+2i)+(3-2i)(3+2i)),\\ \\P(x)=(x-5)(x^2-6x+9-4i^2),\\ \\P(x)=(x-5)(x^2-6x+9+4),\\ \\P(x)=(x-5)(x^2-6x+13),\\ \\P(x)=x^3-11x^2+43x-65.$

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We're given the ODE,

y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) - (9t + 6)

(where I denote exp(x) = eˣ )

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y'' + 4y' + 3y = 0

has characteristic equation

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with roots at r = -1 and r = -3, so the characteristic solution is

y = C₁ exp(-t ) + C₂ exp(-3t )

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y₁ = (at ² + bt + c) exp(-t )

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Compute the derivatives of y₁ :

y₁ = (at ² + bt + c) exp(-t )

y₁' = (2at + b) exp(-t ) - (at ² + bt + c) exp(-t )

… = (-at ² + (2a - b) t + b - c) exp(-t )

y₁'' = (-2at + 2a - b) exp(-t ) - (-at ² + (2a - b) t + b - c) exp(-t )

… = (at ² + (b - 4a) t + 2a - 2b + c) exp(-t )

Substitute them into the ODE [1] to get

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(Note that we don't find out anything about c, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-t ) anyway.)

• y'' + 4y' + 3y = -(9t + 6) … … … [2]

Compute the derivatives of y₂ :

y₂ = at + b

y₂' = a

y₂'' = 0

Substitute these into [2] :

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Use the initial conditions y (0) = 2 and y' (0) = 2 to solve for C₁ and C₂ :

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→ C₁ + 3C₂ = -4 … … … [4]

Solve equations [3] and [4] to get C₁ = 2 and C₂ = -2. Then the particular solution to the initial value problem is

y(t) = -2 exp(-3t ) + (2t ² + t + 2) exp(-t ) - 3t + 2

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