Expert answers to "Consider the following linear second order homogeneous differential..."

ideklaraz7xz

Answered question

2022-03-24

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

\frac{d^{2} y (t)}{{d t}^{2}} + \frac{d y (t)}{d t} - 2 y (t) = 0, y (0) = - 1 \frac{d y (0)}{d t} = 5

Solve the differential equation.

haiguetenteme7zyu

Beginner2022-03-25Added 13 answers

Solution:
Given:

\frac{d^{2} y (t)}{{d t}^{2}} + \frac{d y (t)}{d t} - 2 y (t) = 0

, (1)

y (0) = - 1

and

y^{'} (0) = 5

Auxiliary equation is

m^{2} + m - 2 = 0

\Rightarrow m^{2} + 2 m - m - 2 = 0

m (m + 2) - 1 (m + 2) = 0

(m + 2) (m - 1) = 0

\Rightarrow m_{1} = - 2

and

m_{2} = 1

are the roots of auxiliary equation. Then general solution will be

y (t) = c_{1} e^{2 t} + c_{2} e^{t}

(2)
where

c_{1}

and

c_{2}

are constant derivative of (2) with respect to t, we get

\frac{d}{d t} y (t) = \frac{d}{d t} [c_{1} e^{- 2 t} + c_{2} e^{t}]

y^{'} (t) = c_{1} \cdot \frac{d}{d t} \cdot e^{- 2 t} + c_{2} \cdot \frac{d}{d t} e^{t}

y^{'} (t) = - 2 c_{1} e^{- 2 t} + c_{2} e^{t}

(3)

y (0) = - 1

then from (2)

c_{1} \cdot e^{0} + c_{2} \cdot e^{0} = - 1

\Rightarrow c_{1} + c_{2} = - 1

(4)
Also

y^{'} (0) = 5

. Then from (3)

- 2 c_{1} \cdot e^{0} + c_{2} \cdot e^{0} = 5

\Rightarrow - 2 c_{1} + c_{2} = 5

(5)
Solving equation (4) and (5) we get

c_{2} = 1

and

c_{1} = - 2

using these value of

c_{1}

and

c_{2}

in equation (2) we get

y (t) = - 2 e^{2 t} + e^{t}

Jeffrey Jordon

Expert2022-03-31Added 2605 answers

Answer is given below (on video)

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