Answered: "Consider the following linear second-order homogeneous differential equation..."

Coradossi7xod

Answered question

2022-03-25

Consider the following linear second-order homogeneous differential equation with constant coefficients and two initial conditions

\frac{d^{2} y (t)}{{d t}^{2}} - \frac{1}{3} \cdot \frac{d y (t)}{d t} - \frac{2}{9} \cdot y (t) = 0, y (0) = 0, \frac{d y (0)}{d t} = - 3

Korbin Ochoa

Beginner2022-03-26Added 11 answers

Solution:

\frac{d^{2} y (t)}{{d t}^{2}} - \frac{1}{3} \cdot \frac{d y (t)}{d t} - \frac{2}{9} \cdot y (t) = 0

y (0) = 0, \frac{d y (0)}{d t} = - 3

Auxiliary equation:

m^{2} - \frac{1}{3} m - \frac{2}{9} = 0

9 m^{2} - 3 m - 2 = 0

(m - \frac{2}{3}) (m + \frac{1}{3}) = 0

(factor form)

m = \frac{2}{3}, \frac{- 1}{3}

General solution is

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

(1)
Apply intral conditions

y (0) = 0

\Rightarrow 0 = c_{1} e^{0} + c_{2} e^{0} \Rightarrow c_{1} + c_{2} = 0

(2)
and from (1)

\frac{d y}{d t} (t) = \frac{2}{3} c_{1} e^{\frac{2}{3} \cdot t} - \frac{1}{3} c_{2} e^{- \frac{1}{3} \cdot t}

Apply

\frac{d y (0)}{d t} = - 3

\Rightarrow - 3 = \frac{2}{3} c_{1} e^{0} - \frac{1}{3} c_{2} e^{0} \Rightarrow - 9 = 2 c_{1} - c_{2}

(3)
and equation (2) and (3)

2 c_{1} - c_{2} + c_{1} + c_{2} = - 9

3 c_{1} = - 9 \Rightarrow c_{1} = - 3

By equation (2)

c_{2} = 3

Plug value of

c_{1} = - 3

and

c_{2} = 3

in equation (1) we get

y (t) = - 3 e^{\frac{2}{3} \cdot t} + 3 e^{- \frac{1}{3} \cdot t}

Jeffrey Jordon

Expert2022-03-31Added 2605 answers

Answer is given below (on video)

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