First order linear equationsApplications of ODE

What you will learn today

First order linear equations

Applications of ODE

Differential Equations 1/24

First order linear equationsApplications of ODE

A first order linear differential equation is one that can be put intothe form,

dy

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

where P and Q are continuous functions on a given interval.

Example

y ′ + 1x y = 2 is a first order linear equation.

y ′ + xy2 = 2 is not.

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First order linear equationsApplications of ODE

How to solve this?

Example

y ′ +1

xy = 2

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First order linear equationsApplications of ODE

When an apple accidentally hit Newton’s head, he realizes that ifhe multiplies both sides of the equation by an extra factor x , theequation becomes

xy ′ + y = 2x

(...?)

Differential Equations 4/24

First order linear equationsApplications of ODE

Even more accidentally, he realizes that (xy)′ = xy ′ + y , therefore

d(xy)

dx= 2x

( or substitute z = xy if you like.)Integrate both sides, have∫

d(xy) =

∫2xdx

xy = x2 + C

y = x +C

x

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First order linear equationsApplications of ODE

Wait a minuet...How can we make these ”accidental observations”into a more general rule?Let’s see what happened in our example:Step 1, given the equation

y ′ + P(x)y = Q(x)

we would like to multiply an extra function I (x) (called integratingfactor) on both sides, s.t. the LHS becomes the derivative of somenew function, call it z(x).Step 2, solve the new equation

z ′ = I (x)Q(x)

Step 3, from z(x) get back y(x).

Differential Equations 6/24

First order linear equationsApplications of ODE

Guess: Could step 1 always be achieved for a FOLODE?After multiplying I (x), LHS becomes I (x)y ′ + I (x)P(x)y .Daringly want z(x) = J(x)y , z ′ = J(x)y ′ + J ′(x)y .Compare, we let J(x) = I (x), J ′(x) = I (x)P(x).Therefore I ′(x) = I (x)P(x), which is a separable ODE for I (x)itself!

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First order linear equationsApplications of ODE

Let’s solve for I (x) first.

I (x) = Ce∫P(x)dx

For simplicity can pick C=1.

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First order linear equationsApplications of ODE

Going back to the original problem, we have

z =

∫I (x)Q(x)dx

where z = I (x)y .Therefore y = 1

I (x)

∫I (x)Q(x)dx , where we picked

I (x) = e∫P(x)dx .

Differential Equations 9/24

First order linear equationsApplications of ODE

Theorem

For a FOLODE like y ′ + P(x)y = Q(x), the general solution isy = 1

I (x)

∫I (x)Q(x)dx, where I (x) = e

∫P(x)dx .

A sweet lie has been woven when someone gives you an equation

y ′ = P(x)y + Q(x)

. You can not apply the above formula directly! (why?)

Differential Equations 10/24

First order linear equationsApplications of ODE

Example

dy

dx+ 3x2y = 6x2

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First order linear equationsApplications of ODE

Example

x2y ′ + xy = 1, x > 0, y(1) = 2

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First order linear equationsApplications of ODE

Example

y ′ + 2xy = 1

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

Situation:t: timeE(t): voltage(V)I(t): current(A)R: resistance(Ω)L: inductance(H), a constantVoltage drop due to resistor is RI; due to inductor is LdI

dt .

LdI

dt+ RI = E (t)

where I is the unknown function, E (t) is a given function, this is aFOLODE for I .

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

Example

Solve the initial value problem L=4, R=12, E(t)=60, I(0)=0.What is the limiting value of the current?

Differential Equations 15/24

First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

Models for the growth of population,t: time, (variable)P: number of individuals in the population, (unknown function)For simplicity, let’s assume the rate of population growth isproportion to P:

dP

dt= kP

where k > 0 is a constant.

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

Solving, get

P = Cet

where C depends on the initial condition. This is a model forpopulation growth in ideal condition: sufficient food and resources,no epidemics...

-0.5 0.5 1.0 1.5

-10

-5

5

10

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

A more realistic model takes the resource limit into account: k isnot a constant.Rather, say the environmental resource limit is K, as P

K ⇒ 1,k ⇒ 0.i.e.

k = k1(1− P

K)

which is a separable equation. This model was called the logisticdifferential equation. It was proposed by Pierre-Francios Verhulst(Dutch) in the 1840s.

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

From the equation we can see when P = 0 or P = K , dPdt = 0,

therefore the population stays constant. These are calledequilibrium solutions.The general solution to the logistic model is

P(t) =K

1 + Ae−k1t

where A is a constant.

2 4 6 8 10

2000

4000

6000

8000

10 000

12 000

14 000

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

True or false: when there are no wolves, sheep would eat up all thegrass and die. Therefore the existence of wolves is beneficial to thelong term survival of the specie of sheep.

Differential Equations 20/24

First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

S: population of sheep W: wolves k, r , a, b are constants. On anice afternoon, sheep are wandering on the streets of Philly, whatis the chance that any sheep meets any wolf? It is proportional toboth the density of S and W , so under our postulate say it isproportional to SW .We have a coupled system of differential equations:

dS

dt= kS − aSW

dW

dt= −rW + bSW

This predator-prey system is called the Lotka-Volterra equations.Note the time t doesn’t show up on the RHS of either of theabove equations, also called autonomous.

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

Q: find the equilibrium solutions of the system.Eliminate t from our observation:

dS

dW=

kS − aSW

−rW + bSW

The S −W plane (without t) is called the phase plane, thesolution curves on the phase plane are called phase trajectories. Aphase portrait consists of equilibrium points and typical phasetrajectories.Q: In which direction is a point moving in a trajectory? How fast isit moving? Is the equilibrium solution stable? Is it asymptoticallystable?

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

Look at Prob 9 in §10.6

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First order linear equationsApplications of ODE

Electric circuitsPopulation growthPredator-Prey Systems

What you have learnt today:

First order linear equations

Model for electric circuit,

logistic model for population growth,

Lotka-Volterra model for predator-prey system.

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