GEOGG121: MethodsDifferential Equations, MC etcDr. Mathias (Mat) DisneyUCL GeographyOffice: 113, Pearson BuildingTel: 7670 0592Email: mathias.disney@ucl.ac.ukwww.geog.ucl.ac.uk/~mdisney

• Differential equations– Introduction & importance– Types of DE

• Examples• Solving ODEs

– Analytical methods• General solution, particular solutions• Separation of variables, integrating factors, linear operators

– Numerical methods• Euler, Runge-Kutta

Lecture outline

• TextbooksThese are good UG textbooks that have WAY more detail than we need– Boas, M. L., 1985 (2nd ed) Mathematical Methods in the Physical Sciences, Wiley, 793pp.– Riley, K. F., M. Hobson & S. Bence (2006) Mathematical Methods for Physics &

Engineering, 3rd ed., CUP.– Croft, A., Davison, R. & Hargreaves, M. (1996) Engineering Mathematics, 2nd ed., Addison

Wesley.

• Methods, applications– Wainwright, J. and M. Mulligan (eds, 2004) Environmental Modelling: Finding Simplicity in

Complexity, J. Wiley and Sons, Chichester. Lots of examples particularly hydrology, soils, veg, climate. Useful intro. ch 1 on models and methods

– Campbell, G. S. and J. Norman (1998) An Introduction to Environmental Biophysics, Springer NY, 2nd ed. Excellent on applications eg Beer’s Law, heat transport etc.

– Monteith, J. L. and M. H. Unsworth (1990) Principles of Environmental Physics, Edward Arnold. Small, but wide-ranging and superbly written.

• Links– http://www.math.ust.hk/~machas/differential-equations.pdf– http://www.physics.ohio-state.edu/~physedu/mapletutorial/tutorials/diff_eqs/intro.html

Reading material

• What is a differential equation?– General 1st order DEs

– 1st case t is independent variable, x is dependent variable– 2nd case, x is independent variable, y dependent

• Extremely important – Equation relating rate of change of something (y) wrt to

something else (x)– Any dynamic system (undergoing change) may be amenable to

description by differential equations– Being able to formulate & solve is incredibly powerful

Introduction

dydx

= f y, x( )dxdt= f x, t( )

• Velocity– Change of distance x with time t i.e.

• Acceleration– Change of v with t i.e.

• Newton’s 2nd law– Net force on a particle = rate of change of linear momentum (m

constant so…

• Harmonic oscillator– Restoring force F on a system µ displacement (-x) i.e.– So taking these two eqns we have

Examples

F =dp

dt=d mv( )dt

=mdv

dt=ma

F = −kx

md2x

dt2= −kx

v =dx

dt

a =d2x

dt2

• Radioactive decay of unstable nucleus– Random, independent events, so for given sample of N atoms, no. of

decay events –dN in time dt µ N

– So N(t) depends on No (initial N) and rate of decay

• Beer’s Law – attenuation of radiation– For absorption only (no scattering), decreases in intensity (flux

density) of radiation at some distance x into medium, Φ(x) is proportional to x

– Same form as above – will see leads to exponential decay– Radiation in vegetation, clouds etc etc

Examples

−dN

dt∝N

dφ x( )dx

∝−φ x( )

• Compound Interest– How does an investment S(t), change with time, given an annual

interest rate r compounded every time interval Δt, and annual deposit amount k?

– Assuming deposit made after every time interval Δt

– So as Δtè0

Examples

S t +Δt( ) = S t( )+ rΔt( )S t( )+ kΔt

S t +Δt( )− S t( )Δt

=dS

dt= rS + k

http://www.thecreditexaminer.com/five-things-to-know-about-compound-interest-and-savings/http://www.singaporeolevelmaths.com/tag/compound-interest-formula/`

• Population dynamics– Logistic equation (Malthus, Verhulst, Lotka….)– Rate of change of population P with t depends on Po, growth rate r

(birth rate – death rate) & max available population or ‘carrying capacity’ K

– P << K, dP/dt µ rP but as P increases (asymptotically) to K, dP/dtgoes to 0 (competition for resources – one in one out!)

– For constant K, if we set x = P/K then

Examples

dP

dt= rP 1−

P

K

d

dt

P

K= r

P

K1−

P

K

→

dx

dt= rx 1− x( )

http://www.scholarpedia.org/article/Predator-prey_model#Lotka-Volterra_Model

• Population dynamics: II– Lotka-Volterra (predator-prey) equations – Same form, but now two populations x and y, with time –

– y is predator and yt+1 depends on yt AND prey population (x)– x is prey, and xt+1 depends on xt AND y– a, b, c, d – parameters describing relationship of y to x

• More generally can describe– Competition – eg economic modelling– Resources – reaction-diffusion equations

Examples

dx

dt= x a− by( ),

dy

dt= −y c− dx( )

A ‘phase space’ plot – see later on logistic growth, http://www.scholarpedia.org/article/Predator-prey_model#Lotka-Volterra_Model

Examplesdx

dt= x a− by( ),

dy

dt= −y c− dx( )

http://www.evolution-of-ideas.com/homepage/Mathematics/mathsinner/A%20model%20of%20corruption.htm

• Transport: momentum, heat, mass….– Transport usually some constant (proportionality factor) x driving force– Newton’s Law of viscosity for momentum transport

• Shear stress, τ, between fluid layers moving at different speeds - velocity gradient perpendicular to flow, μ = coeff. of viscosity

– Fourier’s Law of heat transport• Heat flux density H in a material is proportional to (-) T gradient and area

perpendicular to gradient through which heat flowing, k = conductivity. In 1D case…

– Fick’s Law of diffusive transport• Flux density F’j of a diffusing substance with molecular diffusivity Dj across density

gradient dρj/dz (j is for different substances that diffuse through air)

Examples

τ = µdudz

H = −k dTdz

!Fj = −Djdρ j

dzSee Campbell and Norman chapter 6

• Analytical, closed form– Exact solution e.g. in terms of elementary functions

such as ex, log x, sin x

• Non-analytical– No simple solution in terms of basic functions– Solution requires numerical methods (iterative) to

solve– Provide an approximate solution, usually as infinite

series

Types: analytical, non-analytical

• Analytical example– Exact solution e.g.

– Solve by integrating both sides

– This is a GENERAL solution• Contains unknown constants

– We usually want a PARTICULAR solution• Constants known• Requires BOUNDARY conditions to be specified

Types: analytical, non-analytical

dxdt= ax

dxx∫ = a dt∫ ln(x)+ c1 = at + c2 x = eat+ c2−c1( ) = eate c2−c1( )

• Particular solution?– BOUNDARY conditions e.g. set t = 0 to get c1, 2 i.e.

– So x0 is the initial value and we have

– Exponential model ALWAYS when dx/dt µ x• If a>0 == growth; if a < 0 == decay• Population: a = growth rate i.e. (births-deaths)• Beer’s Law: a = attenuation coeff. (amount x absorp. per

unit mass)• Radioactive decay: a = decay rate

Types: analytical, non-analytical

x(t) = x0eat

x(t = 0) = e c2−c1( ) = x0

• Analytical: population growth/decay exampleTypes: analytical, non-analytical

dPdt

= (b− d)P P t( ) = P0eb−d( )t

Log scale – obviously linear….

• ODE (ordinary DE)– Contains only ordinary derivatives

• PDE (partial DE)– Contains partial derivatives – usually case when

depends on 2 or more independent variables– E.g. wave equation: displacement u, as function of

time, t and position x

Types: ODEs, PDEs

d2y

dx2+dy

dx= x

∂2u

∂t2= c

2 ∂2u

∂x2

• ODE (ordinary DE)– Contains only ordinary derivatives (no partials)– Can be of different order

• Order of highest derivative

Types: Order

md2x

dt2= −kx

d2y

dx2+dy

dx= x

dx

dt= xt( )

5

2nd 2nd 1st

• ODE (ordinary DE)– Can further subdivide into different degree

• Degree (power) to which highest order derivative raised

Types: Order -> Degree

dy

dt

3

+ y = sin tdx

dt+ x

2=1

d2y

dx2

2

+dy

dx

3

= 3

1st order3rd degree

1st order1st degree

2nd order2nd degree

• ODE (ordinary DE)– Linear or non-linear?

• Linear if dependent variable and all its derivatives occur only to the first power, otherwise, non-linear

• Product of terms with dependent variable == non-linear• Functions sin, cos, exp, ln also non-linear

Types: Linearity

sin xdy

dx+ y = x

d2y

dx2+ y

2= 0

Linear Non-lineary2 term

Non-linearsin y term

dy

dx+ sin y = 0y

dy

dx= x

2

Non-lineary dy/dx

• General solution– Often many solutions can satisfy a differential eqn– General solution includes all these e.g.– Verify that y = Cex is a solution of dy/dx = y, C is any constant– So

– And for all values of x, and eqn is satisfied for any C– C is arbitrary constant, vary it and get all possible solutions– So in fact y = Cex is the general solution of dy/dx = y

Solving

dy

dx=Ce

x

dy

dx= y

• But for a particular solution– We must specify boundary conditions– Eg if at x = 0, we know y = 4 then from general solution– 4 = Ce0 so C = 4 and – is the particular solution of dy/dx = y that satisfies the

condition that y(0) = 4– Can be more than one constant in general solution– For particular solution number of given independent conditions

MUST be same as number of constants

Solving

y = 4ex

• Analytical: Beer’s Law - attenuation– k is extinction coefficient – absorptivity per unit depth, z (m-1)

– E.g. attenuation through atmosphere, where path length (z) µ1/cos(θsun), θsun is the solar zenith angle

– Take logs:– Plot z against ln(ϕ), slope is k, intercept is ϕ0 i.e. solar radiation

with no attenuation (top of atmos. – solar constant)

– [NB taking logs v powerful – always linearise if you can!]

Types: analytical, non-analytical

dφ z( )dz

= −kφ z( ) φ z( ) = φ0e−kz

lnφ z( ) = lnφ0 − kz

• One point conditions– We saw as general solution of– Need 2 conditions to get particular solution

• May be at a single point e.g. x = 0, y = 0 and dy/dx = 1• So and solution becomes• Now apply second condition i.e. dy/dx = 1 when x = 0 so differentiate

– Particular solution is then

Initial & boundary conditions

d2y

dx2+ y = 0y = Acos x +Bsin x

0 = Acos0+Bsin0 = A y = Bsin x

dy

dx= Bcos x 1= Bcos0 = B

y = sin x

• Verify that satisfies

• Verify that is a solution of– (2nd order, 1st degree, linear)

Solving: examplesd2y

dx2+ y = 0y = Acos x +Bsin x

x = t2+ A ln t +B t

d2x

dt2+dx

dt= 4t

• Two point conditions– Again consider– Solution satisfying y = 0 when x = 0 AND y = 1 when x = 3π/2– So apply first condition to general solution – i.e. and solution is – Applying second condition we see

– And B = -1, so the particular solution is

– If solution required over interval a ≤ x ≤ b and conditions given at both ends, these are boundary conditions (boundary value problem)

– Solution subject to initial conditions = initial value problem

Initial & boundary conditions

d2y

dx2+ y = 0

0 = Acos0+Bsin0 = A

y =1= Bsin 3π2

y = Acos x +Bsin x

y = Bsin x

y = −sin x

• We have considered simple cases so far– Where and so

• What about cases with ind. & dep. variables on RHS?– E.g.

• Important class of separable equations. Div by g(y) to solve

– And then integrate both sides wrt x

Separation of variables

dydx

= f x( ) y = f x( )dx∫

dydx

= f x( )g y( )

1g y( )

dydx

= f x( )

1g y( )

dydx∫ dx = 1

g y( )∫ = f x( )∫ dx

• Equation is now separated & if we can integ. we have y in terms of x– Eg where and

– So multiply both sides by y to give and then integrate both sides wrt x

– i.e. and so and

– If we define D = 2C then

Separation of variables

y∫ dy = e−x dx∫

dydx

=e−x

yf x( ) = e−x g y( ) = 1

y

y dydx

= e−x

y2

2= −e−x +C y2 = −2e−x + 2C

y = ± D− 2e−x

Eg See Croft, Davison, Hargreaves section 18, orhttp://www.cse.salford.ac.uk/profiles/gsmcdonald/H-Tutorials/ordinary-differential-equations-separation-variables.pdfhttp://en.wikipedia.org/wiki/Separation_of_variables

• Form– Where p(x), q(x), r(x) and f(x) are fns of x only– This is inhomogeneous (dep on y)– Related homogeneous form ignoring term independent of y

– Use shorthand L{y} when referring to general linear diff. eqn to stand for all terms involving y or its derivatives. From above

– for inhomogeneous general case– And for general homogenous case

– Eg if then where

2nd order linear equationsp x( ) d

2ydx2

+ q x( ) dydx+ r x( ) y = f x( )

p x( ) d2ydx2

+ q x( ) dydx+ r x( ) y = 0

L y{ }= f x( )L y{ }= 0

d 2ydx2

− 4y = x3 L y{ }=d 2ydx2

− 4yL y{ }= x3

• DEs with two or more dependent variables– Particularly important for motion (in 2 or 3D), where eg position

(x, y, z) varying with time t

• Key example of wave equation– Eg in 1D where displacement u depends on time and position– For speed c, satisfies

– Show is a solution of

– Calculate partial derivatives of u(x, t) wrt to x, then t i.e.

Partial differential equations

∂2u∂t2

= c2 ∂2u∂x2

∂2u∂t2

= 4 ∂2u∂x2

u x, t( ) = sin x + 2t( )

∂u∂x

= cos x + 2t( ) ∂u∂t= 2cos x + 2t( )

– Now 2nd partial derivatives of u(x, t) wrt to x, then t i.e.

– So now

– More generally we can express the periodic solutions as (remembering trig identities)

– and

– Where k is the wave vector (2π/λ); ω is the angular frequency (rads s-1) = 2π/T for period T;

Partial differential equations

∂2u∂x2

= −sin x + 2t( ) ∂2u∂t2

= −4sin x + 2t( )∂2u∂t2

= −4sin x + 2t( ) = 4 −sin x + 2t( )#$ %&= 4∂2u∂x2

http://en.wikipedia.org/wiki/List_of_trigonometric_identitieshttp://www.physics.usu.edu/riffe/3750/Lecture%2018.pdfhttp://en.wikipedia.org/wiki/Wave_vector

u+ x, t( ) = Aei kx−ωt( ) u− x, t( ) = Bei kx+ωt( )

• In 3D?– Just consider y and z also, so for q(x, y, z, t)

• Some v. important linear differential operators– Del (gradient operator)

– Del squared (Laplacian)

• Lead to eg Maxwell’s equations

Partial differential equations

∂2q∂t2

= c2 ∂2q∂x2

+∂2q∂y2

+∂2q∂z2

"

#$

%

&'

http://www.physics.usu.edu/riffe/3750/Lecture%2018.pdf

q+kxkykx x, y, z, t( ) = Aei kxx+kyy+kzz−ωt( )

q−kxkykx x, y, z, t( ) = Aei kxx+kyy+kzz+ωt( )

∇f x, y, z( ) = ∂f∂xx + ∂f

∂yy+ ∂f

∂zz

∇⋅∇f x, y, z( ) = Δ x, y, z( ) = ∂2 f∂x2

x + ∂2 f∂y2

y+ ∂2 f∂z2

z

• Euler’s Method– Consider 1st order eqn with initial cond. y(x0) = y0

– Find an approx. solution yn at equally spaced discrete values (steps) of x, xn

– Euler’s method == find gradient at x = x0 i.e.– Tangent line approximation

SOLVING: Numerical approachesdydx

= f x, y( )

dydx x=x0

= f x0, y0( )

0 x0 x1 x

True solution

Tangent approx.

y0

y1

y

y(x1)

Croft et al., p495Numerical Recipes in C ch. 16, p710http://apps.nrbook.com/c/index.htmlhttp://en.wikipedia.org/wiki/Numerical_ordinary_differential_equations

• Euler’s Method– True soln passes thru (x0, y0) with gradient f(x0, y0) at that point– Straight line (y = mx + c) approx has eqn– This approximates true solution but only near (x0, y0), so only

extend it short dist. h along x axis to x = x1

– Here, y = y1 and – Since h = x1-x0 we see– Can then find y1, and we then know (x1, y1)…..rinse, repeat….

SOLVING: Numerical approaches

y = y0 + x − x0( ) f x0, y0( )

y1 = y0 + x1 − x0( ) f x0, y0( )

0 x0 x1 x

True solution

Tangent approx.

y0

y1

yy(x1)

y1 = y0 + hf x0, y0( )

yi+1 = yi + hf xi, yi( )

Generate series of values iterativelyAccuracy depends on h

Croft et al., p495Numerical Recipes in C ch. 16, p710http://apps.nrbook.com/c/index.html

• Euler’s Method: example– Use Euler’s method with h = 0.25 to obtain numerical soln. of

with y(0) = 2, giving approx. values of y for 0 ≤ x ≤ 1

– Need y1-4 over x1 = 0.25, x2 = 0.5, x3 = 0.75, x4 = 1.0 say, so– with x0 = 0 y0 = 2– And

SOLVING: Numerical approaches

dydx

= −xy2

yi+1 = yi + hf xi, yi( )

yi+1 = yi + 0.25 −xiyi2( )

y1 = 2− 0.25 0( ) 22( ) = 2.000y2 = 2− 0.25 0.25( ) 22( ) =1.750y3 =1.750− 0.25 0.5( ) 1.7502( ) =1.367y4 =1.367− 0.25 0.75( ) 1.3672( ) =1.017

Exercise: this can be solved ANALYTICALLY via separation of variables. What is the difference to the approx. solution?

NB There are more accurate variants of Euler’s method..

• Runge-Kutta methods (4th order here….)– Family of methods for solving DEs (Euler methods are subset)– Iterative, starting from yi, no functions other than f(x,y) needed– No extra differentiation or additional starting values needed– BUT f(x, y) is evaluated several times for each step

– Solve subject to y = y0 when x = x0, use

– where

SOLVING: Numerical approaches

Croft et al., p502Rile et al. p1026Numerical Recipes in C ch. 16, p710http://apps.nrbook.com/c/index.html

http://en.wikipedia.org/wiki/Numerical_ordinary_differential_equationshttp://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods

dydx

= f x, y( )

k1 = f xi, yi( ),k2 = f xi +h2, yi +

h2k1

!

"#

$

%&,k3 = f xi +

h2, yi +

h2k2

!

"#

$

%&,k4 = f xi + h, yi + hk3( )

yi+1 = yi +h6k1 + 2k2 + 2k3 + k4( )

Euler

• Runge-Kutta example– As before, but now use R-K with h = 0.25 to obtain numerical

soln. of with y(0) = 2, giving approx. values of y for 0 ≤ x ≤ 1

– So for i = 0, first iteration requires

– And finally

– Repeat! c.f. 2 from Euler, and 1.8824 from analytical

SOLVING: Numerical approaches

dydx

= −xy2

k1 = f x0, y0( ) = − 0( ) 2( )2 = 0

k2 = f 0.125, 2( ) = − 0.125( ) 2( )2 = −0.5

k3 = f 0.125, 2+0.252

−0.5( )"

#$

%

&'= f 0.125,1.9375( ) = −0.125 1.9375( )2 = −0.4692

k4 = f 0.25, 2+ 0.25 −0.4692( )( ) = f 0.25,1.8827( ) = −0.25 1.9375( )2 = −0.8861

y1 = 2+0.256

0+ 2 −0.5( )+ 2 −0.4692( )+ −0.8861( )( ) =1.8823

• Brute force method(s) for integration / parameter estimation / sampling– Powerful BUT essentially last resort as involves random

sampling of parameter space– Time consuming – more samples gives better approximation– Errors tend to reduce as 1/N1/2

• N = 100 -> error down by 10; N = 1000000 -> error down by 1000

– Fast computers can solve complex problems

• Applications:– Numerical integration (eg radiative transfer eqn), Bayesian

inference, computational physics, sensitivity analysis etc etc

Very brief intro to Monte Carlo

Numerical Recipes in C ch. 7, p304http://apps.nrbook.com/c/index.htmlhttp://en.wikipedia.org/wiki/Monte_Carlo_method

http://en.wikipedia.org/wiki/Monte_Carlo_integration

• Pick N random points in a multidimensional volume V, x1, x2, …. xN

• MC integration approximates integral of function f over volume V as

• Where and

• +/- term is 1SD error – falls of as 1/N1/2

Basics: MC integration

Fromhttp://apps.nrbook.com/c/index.html

Choose random points in AIntegral is fraction of points under curve x A

f dV ≈V f∫ ±Vf2 − f

2

N

f ≡1

Nf xi( )

i=1

N

∑ f2

≡1

Nf2xi( )

i=1

N

∑

• Why not choose a grid? Error falls as N-1 (quadrature approach)• BUT we need to choose grid spacing. For random we sample until

we have ‘good enough’ approximation• Is there a middle ground? Pick points sort of at random BUT in

such a way as to fill space more quickly (avoid local clustering)?• Yes – quasi-random sampling:

– Space filling: i.e. “maximally avoiding of each other”

Basics: MC integration

Sobol method v pseudorandom: 1000 pointsFROM: http://en.wikipedia.org/wiki/Low-discrepancy_sequence

• Differential equations– Describe dynamic systems – wide range of examples, particularly

motion, population, decay (radiation – Beer’s Law, mass –radioactivity)

• Types– Analytical, closed form solution, simple functions– Non-analytical: no simple solution, approximations?– ODEs, PDEs– Order: highest power of derivative

• Degree: power to which highest order derivative is raised

– Linear/non:• Linear if dependent variable and all its derivatives occur only to the first

power, otherwise, non-linear

Summary

• Solving– Analytical methods?

• Find general solution by integrating, leaves constants of integration• To find a particular solution: need boundary conditions (initial, ….) • Integrating factors, linear operators

– Numerical methods?• Euler, Runge-Kutta – find approx. solution for discrete points

• Monte Carlo methods– Very useful brute force numerical approach to integration, parameter

estimation, sampling– If all else fails, guess…..

Summary

END

• Radioactive decay– Random, independent events, so for given sample of N atoms, no. of

decay events –dN in time dt µ N so

– Where λ is decay constant (analogous to Beer’s Law k) units 1/t– Solve as for Beer’s Law case so – i.e. N(t) depends on No (initial N) and rate of decay– λ often represented as 1/tau, where tau is time constant – mean

lifetime of decaying atoms– Half life (t=T1/2) = time taken to decay to half initial N i.e. N0/2– Express T1/2 in terms of tau

Example

dN

dt∝−N

dN

dt= −λN

N t( ) = N0e−λt

• Radioactive decay– EG: 14C has half-life of 5730 years & decay rate = 14 per minute per

gram of natural C– How old is a sample with a decay rate of 4 per minute per gram?– A: N/N0 = 4/14 = 0.286– From prev., tau = T1/2/ln2 = 5730/ln2 = 8267 yrs– So t = -tau x ln(N/N0) = 10356 yrs

Example

• General solution of is given by

• Find particular solution satisfies x = 3 and dx/dt = 5 when t =0

• Resistor (R) capacitor (L) circuit (p458, Croft et al), with current flow i(t) described by

• Use integrating factor to find i(t)….approach: re-write as

Exercisesd 2xdt2

−3dxdt+ 2x = 0 x = Aet +Be2t

iR+ L didt= t, t ≥ 0, i 0( ) = 0

didt+RLi = t

L

• Show that the analytical solution of with y(x=0)=2 is

• Compare values from x = 0 to 1 with approx. solution obtained by Euler’s method

•

Exercisesdydx

= −xy2

y = 2x2 +1

• For equations of form– Where P(x) and Q(x) are first order linear functions of x, we can

multiply by some (as yet unknown) function of x, μ(x)

– But in such a way that LHS can be written as– And then

– Which is said to be exact, with μ(x) as the integrating factor– Why is this useful?

Using an integrating factordydx+P x( ) y =Q x( )

ddx

µ x( ) y( )

Eg See Croft, Davison, Hargreaves section 18, orhttp://www.cse.salford.ac.uk/profiles/gsmcdonald/H-Tutorials/ordinary-differential-equations-integrating-factor.pdfhttp://en.wikipedia.org/wiki/Integrating_factor

µ x( ) dydx+µ x( )P x( ) y = µ x( )Q x( )

ddx

µ x( ) y( ) = µ x( )Q x( )

• Because it follows that

• And if we can evaluate the integral, we can determine y

• So as above, we want

• Use product rule i.e. and so, from above

• and by inspection we can see that

• This is separable (hurrah!) i.e.

Using an integrating factor

ddx

µy( ) = µ dydx+µPy

http://en.wikipedia.org/wiki/Product_rulehttp://www.cse.salford.ac.uk/profiles/gsmcdonald/H-Tutorials/ordinary-differential-equations-integrating-factor.pdf

µ x( ) y = µ x( )Q x( )∫ dxddx

µ x( ) y( ) = µ x( )Q x( )

ddx

µy( ) = µ dydx+dµdx

y

µdydx+dµdx

y = µ dydx+µPy

dµdx

= µP

dµµ

∫ = Pdx∫

• And we see that (-lnK is const. of integ.)• And so

• We can choose K = 1 (as we are multiplying all terms in equation by integ. factor it is irrelevant), so

– Integrating factor for is given by

– And solution is given by

Using an integrating factor

http://en.wikipedia.org/wiki/Product_rulehttp://www.cse.salford.ac.uk/profiles/gsmcdonald/H-Tutorials/ordinary-differential-equations-integrating-factor.pdf

µ x( ) = eP x( )dx∫

lnµ − lnK = Pdx∫ln µK= Pdx∫

dydx+P x( ) y =Q x( )

µ x( ) y = µ x( )Q x( )∫ dx

• Solve

– From previous we see that and

– Using the formula above

– And we know the solution is given by

– So , as

Using an integrating factor: example

µ x( ) = eP x( )dx∫ = e

1xdx∫= eln x = x

dydx+yx=1 dy

dx+P x( ) y =Q x( )

P x( ) = 1x

Q x( ) =1

xy = x dx∫ =x2

2+C

µ x( ) y = µ x( )Q x( )∫ dx

Q x( ) =1

• When L{y} = f(x) is a linear differential equation, L is a linear differential operator– Any linear operator L carries out an operation on functions f1

and f2 as follows1.2. where a is a constant3. where a, b are constants

– Example: if show that– and

Linear operators

L f1 + f2{ }= L f1{ }+ L f2{ }L af1{ }= aL f1{ }L af1 + bf2{ }= aL f1{ }+ bL f1{ }

L y{ }=d 2ydx2

+3x dydx− 2y L y1 + y2{ }=

L y1{ }+ L y2{ }L ay{ }= aL y{ }

• Note that L{y} = f(x) is a linear diff. eqn so L is a linear diff operator• So

– we see

– And rearrange:

– & because differentiation is a linear operator we can now see

• For the second case

• So

Linear operators

=d 2y1dx2

+3x dy1dx

− 2y1 +d 2y2dx2

+3x dy2dx

− 2y2

L y1 + y2{ }=d 2

dx2y1 + y2( )+3x d

dxy1 + y2( )− 2 y1 + y2( )

L y1 + y2{ }= L y1{ }+ L y2{ }

L ay{ }=d 2 ay( )dx2

+3xd ay( )dx

− 2 ay( )

L ay{ }=ad 2ydx2

+3ax dydx− 2ay = a d 2y

dx2+3x dy

dx− 2y

"

#$

%

&'= aL y{ }