Expert Answer to "A linear differential equationx″+2cx′+(2cos⁡h2t−1)x=0where c is a constant."

Name: A linear differential equationx″+2cx′+(2cos⁡h2t−1)x=0where c is a constant.
Uploaded: 2022-02-23T17:22:57+00:00
Description: A linear differential equationx″+2cx′+(2cos⁡h2t−1)x=0where c is a constant.

Dowqueuestbew1j

Answered question

2022-01-21

A linear differential equation

x^{″} + 2 c x^{'} + (\frac{2}{{\cos h}^{2} t} - 1) x = 0

where c is a constant.

Answer & Explanation

sonorous9n

Beginner2022-01-21Added 34 answers

When

c = 0

we can guess that

x_{1} = \frac{1}{\cos h t}

is a solution. Then by inserting

x = x_{1} \cdot u

into equation it simplifies to:

u^{″} - 2 \tan h t u^{'} = 0

From which we conclude that:

x (t) = \frac{C_{1}}{\cos h t} + C_{2} (\sin h t + \frac{t}{\cos h t})

For more general case when

c \neq 0

we use substitution

x = w e^{- c t}

. The equation transforms to:

w^{″} = (1 + c^{2} - \frac{2}{{\cos h}^{2} t}) w

(1)
the trick is to use Darboux theorem. Let's say we have two functions y and z that are solutions to:

y^{″} = p (t) y, z^{″} = (p (t) - λ) z

Then the function:

\overset{―}{y} = y^{'} - \frac{y z^{'}}{z}

Satisfies:

\overset{―}{y}^{″} = (p - 2 \frac{d}{d t} (\frac{z^{'}}{z})) \overset{―}{y}

(2)
Taking

z = \cos h t and p = 1 + c^{2}

equation (2) turns out to be (1).
We solve:

y^{″} = (1 + c^{2}) y \Rightarrow y = C_{1} e^{\sqrt{1 + c^{2} t}} + C_{2} e^{- \sqrt{1 + c^{2} t}}

then we plug in for

\overset{―}{y} = w

and x which yields the final result:

x = C_{1} (\sqrt{1 + c^{2}} - \tan h t) e^{\sqrt{1 + c^{2}} t - c t}

+ C_{2} (\sqrt{1 + c^{2}} + \tan h t) e^{- \sqrt{1 + c^{2}} t - c t}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Didn't find what you were looking for?