"Find the general solution to the following differential..." was answered by our experts

Vikolers6

Answered question

2021-12-27

Find the general solution to the following differential equations.

y 6 y^{'} + 13 y = 0

Answer & Explanation

alexandrebaud43

Beginner2021-12-28Added 36 answers

Given:

y 6 y^{'} + 13 y = 0

y 6 y^{'} + 13 y = 0

A second order linear, homogenous ODE has the form

a y b y^{'} + c y = 0

For an equation

a y b y^{'} + c y = 0

, assume a solution of the form

e^{λ t}

Rewrite the equation with

y = e^{λ t}

(e^{λ t}) 6 {(e^{λ t})}^{'} + 13 e^{λ t} = 0

e^{λ t} (y^{2} + 6 y + 13) = 0

y^{2} + 6 y + 13 = 0

Roots of the quadratic equation in the form

a x^{2} + b x + c = 0

x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Here,

a = 1, b = 6 and c = 13

y_{1, 2} = \frac{- 6 \pm \sqrt{6^{2} - 4 \times 1 \times 13}}{2 \times 1}

y_{1, 2} = \frac{- 6 \pm \sqrt{36 - 52}}{2}

y_{1, 2} = - 3 - 2 i, - 3 + 2 i

For two complex roots

y_{1} \neq y_{2}

, where

y_{1} = - 3 - 2 i, y_{2} = - 3 + 2 i

So, the general solution takes the form:

y = e^{α t} (c_{1} \cos (β t) + c_{2} \sin (β t))

y = e^{- 3 t} (c_{1} \cos (2 t) + c_{2} \sin (2 t))

xandir307dc

Beginner2021-12-29Added 35 answers

y = e^{k x}

k^{2} + 6 k + 13 = 0

D = b^{2} - 4 a c = 6^{2} - 4 \cdot 1 \cdot 13 = - 16

\sqrt{D} = \pm 4 i

k_{1, 2} = - 3 \pm 2 i

Answer:

y = e^{- 3 x} (C_{1} \cos 2 x + C_{2} \sin 2 x)

karton

Expert2022-01-10Added 613 answers

We were given the next equation
$y^{″} + 6 y^{'} + 13 y = 0 (1)$
The roots of the characteristic equation
$λ^{2} + 6 λ + 13 = 0 (2)$
are $λ_{1} = - 3 + 2 i, λ_{2} = - 3 - 2 i$
The general solution of Eq. (1) is, by the Theorem 4.3.2, given by
$y (t) = c_{1} e^{μ t} \cos v t + c_{2} e^{μ t} \sin v t$
where $λ_{1, 2} = μ \pm i v$ . In this case, the solution is
$y (t) = c_{1} e^{- 3 t} \cos 2 t + c_{2} e^{- 3 t} \sin 2 t$

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