Lecture 2Differential equations

Ing. Jaroslav Jíra, CSc.

Physics for informatics

Basic division of differential equations

According to type of derivation

Ordinary Differential Equations – ODEThey contain a function of one independent variable and its derivatives.

Example:

Partial Differential Equations - PDE They contain unknown multivariable functions and their partial derivatives.

Example:

xyyyorxydx

dy

dx

yd 22

2

2

),,,(

12

2

22

2

2

2

2

2

tzyxuwhere

dt

u

cdz

u

dy

u

dx

u

Types of Ordinary Differential Equations

First order differential equationsThe highest derivative of the function y of independent variable they contain is one.

Example:

Higher order differential equations They contain at least second derivative of the function y

Example:

xyy 2

xyyy 23

General definition of an ODE

0),....,,,(),....,,,( )()()1( nnn yyyyxForyyyyyxF

Types of Ordinary Differential Equations

Linear differential equationsThe function y appears just linearly here. There are no powers of the function y and its derivatives, there are no products of the funciton y and its derivatives. There are also no functions of the function y like sin(y), exp(y) etc.

Example:

Nonlinear differential equations If any of conditions previously defined for the linear DE is not met, then we talk about nonlinear DE.

Example:

xyyyy 23

000)sin( 2 yyoryyyoryy

Types of Ordinary Differential Equations

Homogeneous differential equationsThere is no constant or function of x on the right side of the equation.

Example:

Inhomogenous differential equations Exact opposite to the homogenous DE.

Example:

02 yyy

xyyoryy 3242

Types of Ordinary Differential Equations

Differential equations with constant coefficientsThe function y and all its derivatives are mutiplied just by constants.

Example:

Differential equations with variable coefficients The function y and its derivatives are mutiplied either by constants or by functions of x.

Example:

xyyyy 345

xyyeyx x 2)1(3

How to solve differential equations

The following methods will be mentioned subsequently

Separation of variables – applicable to homogeneous first order ODEs and to specific types of inhomogeneous first order ODEs.

Characteristic equation – applicable to homogeneous first order and higher order linear ODEs with constant coefficients

Integrating factor – applicable to inhomogeneous first order and higher order linear ODEs. This method usually completes the solution of previous two methods for inhomogeneous equations.

The most simple differential equation: )(' xfy

Integrating the last equation Cdxxfy )(

Example: xy 2'

dx

dyy 'where

CxCdxxy 22

Where x2+C is general solution of the differential equation

dxxfdy )(

Sometimes an additional condition is given like

3)2( y

that means the function y(x) must pass through a point ]3,2[0 x

123 2 CC

We have obtained a particular solution y(x)=x2 -1

1)( 2 xxy

First order inhomogenous linear differential equation with constant coefficients – separation of variables

?)( xy

)(xfdx

dy

First order homogenous linear differential equation with constant coefficients – separation of variables

0' byayThe general formula for such equation is ?)( xy

Now we integrate both sides of the equation and then we apply an exponential function to it

Cxa

by ln

substituting xa

b

Cey

Where C is a constant resulting from the initial condition

0bydx

dya y

a

b

dx

dy dx

a

b

y

dy

)(ln Cxa

by ee

xa

bC eey

CeC we obtain general solution

First order homogenous linear differential equation with constant coefficients – characteristic equation

0' byayThe general formula for such equation is

xx Ceyorey

?)( xy

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

xey xey 'If then

and the equation changes into

0baafter dividing by the eλx we obtaina

b

0 xx ebea 0)( bae x

the solution isx

a

b

Cey

Where C is a constant resulting from the initial condition

This is the characteristic equation

Example of the first order linear ODE – RC circuit

CR uuu

Find the time dependence of the electric current i(t) in the given circuit.

1.u Ri idt

C

Now we take the first derivative of the last equation with respect to time

011

0 iRCdt

dii

Cdt

diR

characteristic equation is

RCRC

10

1

general solution is

tRCKei1

Constant K can be calculated from initial conditions. We know that

R

uKKe

R

u

R

ui RC

0

)0(

tRCe

R

ui

1

particular solution is

Solution of the RC circuit in the Mathematica

Given values are R=1 kΩ; C=100 μF; u=10 V

First order inhomogenous linear differential equation with constant coefficients – integrating factor

QPyy 'The general formula for such equation is

Where P(x) and Q(x) are continuous functions of variable x.

PdxCeyCPdxy lnln

Having solution of homogeneous part of the equation we can continue by looking for the integrating factor. Instead of constant C we are looking for a function C(x), which satisfies both homogeneous solution and the original differential equation.

In the first step we omit the right side taking Q(x)=0 and than we can solve the homogeneous equation by separation of variables.

Separation of variables gives us: Pdxy

dyPy

dx

dy 0

)( dxQeKey dxPPdx

If C(x) should be a function, then

Substituting for y and y’ into the original equation QPyy

The general solution of this inhomogeneous equation is:

we obtain

and after small adjustment:

PdxexCy )(

PdxPdx PexCexCy )()(

QexPCPexCexC PdxPdxPdx )()()(

QexC Pdx )(

Our integrating factor is: KdxeQxC Pdx )(

Example of the integrating factor solution – LR circuit

LR uuU

Find the time dependence of the electric current i(t) in the given circuit.

Firstly we have to solve homogeneous equation

00 RLRidt

diL

Solution of charact. equation

In the second step we are looking for the integrating factor

tL

R

etKisotK

)()( 11

First derivative of the current

dt

diLRiU

tL

R

eKtiL

R 1)(

tL

Rt

L

R

etL

RtKetK

dt

di )()()( 11

General solution

Knowing that initial current is zero i(0)=0, we can determine K2.

Particular solution of the equation

Substituting for i and di/dt into the original equation gives us

UetRKetL

RtKetKL

tL

Rt

L

Rt

L

R

)())()()(( 111

2111 )()()( KeR

UtKe

L

UtKUetLK

tL

Rt

L

Rt

L

R

tL

Rt

L

R

eKeR

Ui

)( 2

tL

R

eKR

Ui

2

R

UKeK

R

U 2

020

)1(t

L

R

eR

Ui

Solution of the RC circuit in the Mathematica

Given values are R=1 kΩ; L=100 mH; U=10 V

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

Damping constant δ=1 [s-1], angular frequency ω=10 [s-1]

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