SECOND ORDER LINEAR Des WITH CONSTANT COEFFICIENTS

Second order homogenous linear differential equation with constant coefficients

0''' cybyayThe general formula for such equation is

xey To solve this equation we assume the solution in the form of exponential function:

xey xey 'If then

and the equation will change into

02 cba after dividing by the eλx we obtain

02 xxx ecebea

0)( 2 cbae x

We obtained a quadratic characteristic equation. The roots are

xey 2'' and

a

acbb

2

42

12

?)( xy

There exist three types of solutions according to the discriminant D acbD 42

1) If D>0, the roots λ1, λ2 are real and distinct

xx eCeCy 2121

2) If D=0, the roots are real and identical λ12 =λ

xx xeCeCy 21

3) If D<0, the roots are complex conjugate λ1, λ2 where α and ω are real and imaginary parts of the root

i

i

2

1

xixxixxx eKeKeKeKy 212121

)( 21xixix eKeKey

]sin)(cos)[( 2121 xKKixKKey x

formulaEulers

xixe xi sincos

)();( 212211 KKiCKKC

]sincos[)( 21 xCxCexy x

If we substitute

we obtain

Further substitution is sometimes used cos;sin 21 ACAC

]sincoscossin[)( xAxAexy x and then

sincoscossin)sin( considering formula

we finally obtain )sin()( xAexy x

where amplitude A and phase φ are constants which can be obtained from initial conditions and ω is angular frequency.This example leads to an oscillatory motion.

This is general solution in some cases, but …

Example of the second order LDE – a simple harmonic oscillator

Evaluate the displacement x(t) of a body of mass m on a horizontal spring with spring constant k. There are no passive resistances.

xkF If the body is displaced from its equilibrium position (x=0), it experiences a restoring force F, proportional to the displacement x:

From the second Newtons law of motion we know

xmdt

xdmmaF

2

2

0 xm

kxkxxm Characteristic

equation is02

m

k

We have two complex conjugate roots with no real part m

ki12

)sin()( tAetx tThe general solution for our symbols is

No real part of λ means α=0, and omega in our casem

k

)sin()( tAtxThe final general solution of this example is

Answer: the body performs simple harmonic motion with amplitude A and phase φ. We need two initial conditions for determination of these constants.

These conditions can be for example

)cos(2)( ttx

20cos0)0cos(

A

The particular solution is

2)0(0)0( xx

From the first condition

From the second condition

22)2

0sin( AA

)2

sin(2)( ttx

Example 2 of the second order LDE – a damped harmonic oscillator

The basic theory is the same like in case of the simple harmonic oscillator, but this time we take into account also damping.

The damping is represented by the frictional force Ff, which is proportional to the velocity v.

xcdt

dxcvcFf

The total force acting on the body is xckxFkxF f

xmmaF 0 kxxcxmxckxxm

0 xm

kx

m

cx The following substitutions are

commonly used m

c

m

k 2;

02 2 xxx Characteristic equation is 02 22

2222

12 2

442

Solution of the characteristic equation

where δ is damping constant and ω is angular frequency

There are three basic solutions according to the δ and ω.

1) δ>ω. Overdamped oscillator. The roots are real and distinct

222

221

tt eCeCtx 2121)(

2) δ=ω. Critical damping. The roots are real and identical. 12

tt teCeCtx 21)(

3) δ<ω. Underdamped oscillator. The roots are complex conjugate.

'

'

222

221

ii

ii

)'sin()( tAetx t

Damped harmonic oscillator in the Mathematica

All three basic solutions together for ω=10 s-1 Overdamped oscillator, δ=20 s-1

Critically damped oscillator, δ=10 s-1

Underdamped oscillator, δ=1 s-1