section 2.1 : separable equationspeople.math.gatech.edu/~jbeardsley6/chapter2slides.pdfsection 2.1 :...

Section 2.1 : Separable Equations

Chapter 2 : First Order Differential Equations

Math 2552 Differential Equations

Section 2.1 Slide 1

Section 2.1

TopicsWe will cover these topics in this section.

1. Solving first order separable differential equations.

ObjectivesFor the topics covered in this section, students are expected to be able todo the following.

1. Classify differential equations as separable

2. Solve separable first order linear ODEs

Section 2.1 Slide 2

Lecture Organization

� We explore a method to solve separable differential equations inthis section.

� We will

1. define what a separable differential question is2. then solve an ODE3. then discuss the solution method more generally.

� Our approach will make use of the differential, a topic oftencovered in a differential calculus course.

Section 2.1 Slide 3

Definition

A first order differential equation is separable if it can be writtenin the form

dy

dx= f(x, y) = p(x)q(y)

Definition: Separable ODE

Section 2.1 Slide 4

Example

Consider the differential equation

dy

dt= (1− 12t)y2, y = y(t), y(0) =

1

8.

a) Is this differential equation linear?

b) Compute the differential dy = dydt dt.

c) If possible, separate variables on either side of the equation in theprevious step.

d) If possible, integrate and solve for y.

Section 2.1 Slide 5

Section 2.1 Slide 6

General Procedure

This gives us a procedure for solving a separable DE.

1. Compute the differential dy = dydt dt.

2. Separate variables on either side of the equation in the previous step.

3. Integrate and solve for y.

Note that our differential equation need not be linear for us to use thismethod.

Section 2.1 Slide 7

Section 2.2 : Linear Equations: IntegratingFactors



Section 2.2 Slide 8

Section 2.2


1. Solving a first order linear ordinary differential equations using aprocedure that uses an integrating factor


1. Convert differential equations into standard form

2. Classify differential equations as linear

3. Solve first order linear ODEs using an integrating factor

Section 2.2 Slide 9

Lecture Organization

� We explore a method to solve a first order linear ODE in thissection.

� Recall that an ODE is in standard form if it can be written as

dy

dt+ p(t)y(t) = y(t)

� We will solve an equation in standard form on the next slide, andthen discuss the solution method more generally.

Section 2.2 Slide 10

Example 1

Consider the differential equation ty′ + 2y = 4t for t ≥ 0.

1. Is this differential equation separable?

2. Solve the differential equation.


The Integrating Factor

Given a linear first order ODE in standard form,

dy

dt+ p(t)y = g(t) (1)

Multiply by the integrating factor µ(t) = e∫p dt.

µ(t)dy

dt+ p(t)µ(t)y = µ(t)g(t) (2)

µ is constructed so that the left-hand side is a derivative of a product.

d

dt(µ(t)y(t)) = µ(t)

dy

dt+ p(t)µ(t)y (3)

We construct µ so that the left-hand side of (2) is the derivative of µ(t) y.


Procedure

This gives us a procedure for solving a first order linear DE.

dy

dt+ p(t)y(t) = g(t)

1. Convert given DE to standard form (if necessary).

2. Calculate integrating factor µ(t) = e∫p(t) dt

3. Multiply DE by µ, express DE in the form [µ(t)y]′= µg.

4. Integrate

5. Solve for y

We will explore more examples of solving linear equations in the nextsection.


Section 2.3 : Modeling with First OrderEquations




Section 2.3


1. Solving a first order linear ordinary differential equations using aprocedure that uses an integrating factor


1) Construct a differential equation to model a real-world situation.

2) Solve the differential equation so that we can interpret its solutionto characterize a system.

3) Analyze mathematical statements and solutions of differentialequations.


Example 1: Water Tank

A tank initially contains 40 pounds of salt dissolved in 600 gallons ofwater. Starting at time t = 0, water that contains 1/2 pound of salt pergallon is poured into the tank at the rate of 4 gal/min and the mixture isdrained from the tank at the same rate.

a) Construct a differential equation for Q(t), which is the number ofpounds of salt in the tank at time t > 0.

b) Solve the DE to determine an expression for Q(t).

c) After a long period of time, what happens to the concentration ofsalt in the tank?


Mathematical Modeling

The process we used in the previous example roughly followed thisprocess.

1) Construct a differential equation to model a real-world situation.

2) Solve the differential equation so that we can interpret its solutionto characterize a system.

3) Analyze mathematical statements and solutions of differentialequations.

The above process is used throughout this course, and so are captured inthe course learning objectives that are stated in the syllabus.


Example 2: Population Model

The world population in 2018 is roughly 7.6 billion.

a) The world population is increasing at a rate of 1.2% per year. If thegrowth rate remains fixed at 1.2%, how long will it take for thepopulation of the world to reach 20 billion people?

b) Assume the earth cannot support a population beyond 20 billionpeople. If the population growth rate is also proportional to thedifference between how close the world population is to this limitingvalue, what is the expression that gives the world population as afunction of time?


Section 2.4 : Differences Between Linear andNonlinear Equations




Section 2.4


1. Theorems for first order linear and nonlinear IVPs.


1) Characterize first order linear and nonlinear IVPs in terms ofexistence and uniqueness.

2) Determine intervals where a solution to a given first order IVP exists.


Motivation

There are two questions that we exploring in this section.

Given an initial value problem (IVP).

1. existence: does the IVP have a solution, and if so, where?

2. uniqueness: is the solution unique?

We explore these questions for linear and non-linear cases.


A Motivating Example

Consider the IVPdy

dt= 8ty1/5, y(0) = 0

Take a few minutes to answer the following on your own.

a) Is the ODE linear?

b) Solve the IVP to determine an expression for y(t).

c) Is your solution unique? Can you identify another solution to theIVP?

After a few minutes compare your results with someone sitting nearby.


Linear IVP

If p and g are continuous on (α, β), t0 ∈ (α, β), then there is a uniquesolution to the IVP

y′ + p(t)y = g(t), y(t0) = y0

Theorem: Existence and Uniqueness of 1st Order Linear IVP

A proof of this theorem is in the textbook.

Example: Consider the IVP

(9− t2)y′ + 5ty = 3t2, y(−1) = 1

Determine an interval where a solution to the IVP exists.


Nonlinear IVP

If f and ∂f∂y are continuous over α < t < β, and γ < y < δ which

contains the point (t0, y0), then there is a unique solution to the IVP

y′ = f(t, y), y(t0) = y0

on an interval contained in α < t < β.

Theorem: Existence and Uniqueness of 1st Order Nonlinear IVP

Note that:

� A proof of this theorem goes beyond the scope of this course.

� These conditions are sufficient, but not necessary.

� The expression∂f

∂yis a partial derivative.

� Doesdy

dt= 8ty1/5, y(0) = 0, satisfy the conditions of this theorem?


Other Linear and Nonlinear IVP Comparisons

The 1st order ODE y′ + p(t)y = g(t) has the following properties.

1. If the p and g are continuous, there is a general solution containingan arbitrary constant that represents all solutions of the ODE.

2. There is an explicit expression for the ODE.

3. Points where the solution is discontinuous can be found by identifiedwithout solving the ODE: they are identified from the coefficients.

A nonlinear first order ODE does not necessarily have any of the aboveproperties.


Section 2.5 : Autonomous Equations andPopulation Dynamics




Section 2.5


1. Bifurcation points and bifurcation diagrams

2. Concavity


1) Use concavity to sketch solution curves of a DE.

2) Sketch a bifurcation diagram for a first order autonomous DE.


Autonomous DEs

Recall the following.

� an autonomous DE has the formdy

dt= f(y)

� equilibrium solutions of an autonomous DE can be found bylocating roots of f(y) = 0

� we can use equilibrium solutions to sketch solution curves

A more accurate sketch of solution curves can be drawn by usingconcavity.

Note that if dydt = f(y), then by the Chain Rule,

d2y

dt2=

d

dtf (y) =

df

dy

dy

dt

Thus, if y is concave up when:


Example 1

A system obeys the logistic equationdy

dt= ry

(1− y

K

), for y ∈ R,

r > 0, and K > 0.

a) Sketch f vs y, identify and classify the equilibrium points of y.

b) For y ∈ R determine whether y is concave up or concave down.

c) Use the information in parts (a), (b) to sketch a few integral curvesof the DE.


Example 2 (as time permits)

A population obeys the DE y′ = a− y2, for a parameter a ∈ R. Take afew minutes to solve the following on your own.

a) Determine the equilibrium points for any a ∈ R. There are threecases.

b) Sketch the phase lines for each case and classify the critical points.

c) For the case when a > 0, sketch a few solution curves.

d) Sketch the location of the critical point as a function of a in anay-plane.

The sketch in part (d) is known as a bifurcation diagram.


Section 2.6 : Exact Equations and IntegratingFactors




Section 2.6


1. Exact differential equations.


1) Identify whether a given DE is exact.

2) Solve exact DEs.

The use of integrating factors to convert a DE into one that is exact isexplored in the textbook and goes beyond the scope of our course.


Hanging Chain

A chain is hanging off the end of a table.

Photo credit: www.flickr.com/photos/ironpoison photos 69


Hanging Chain

A chain is hanging off the end of a table.

The chain starts to fall.

� x = length of the portion of the chain that is hanging

� v = velocity of the chain

Ignoring resistance, it can be shown that

xv2 − 32x2 + x2vdv

dx= 0, x > 0, v > 0

We wish to solve this DE to obtain v = v(x).

Two questions:

1. how would you classify this DE?

2. hypothesis: what should our solution, v(x), look like?


Hanging Chain - Solution

Our DE is:

xv2 − 32x2 + x2vdv

dx= 0

Or

M +Ndv

dx= 0

It helps to express our DE in differential form:


Hanging Chain - Solution Plot

If we assume that v = 1 when x = 1, then we can solve for the constantof integration and plot the solution to see if the solution matches ourguess of what v(x) should look like.

Let’s plot our solution at desmos.com.


Exact DE

A differential equation

M(x, y) +N(x, y)dy

dx= 0

is exact if and only if

∂M

∂y=

Then there exists a function ψ = ψ(x, y) that satisfies

∂ψ

∂x=M,

∂ψ

∂y= N

Theorem: Exact DE


Solving an Exact DE

In our falling chain example, we had a DE of the form

M(x, y) +N(x, y)dy

dx= 0

We determined that the DE was exact. In other words,

0 =M +Ndy

dx=∂ψ

∂x+∂ψ

∂y

dy

dx=dψ

dx

This is the differential of ψ (x, y(x)).

Finally, by integration:


Summary

Given a first order exact DE

M(x, y) +N(x, y)dy

dx= 0

To solve this exact DE:

1. Determine ψ. One way to do this:

a) integrate M with respect to x, which introduces a function h(y)

ψ =

∫Mdx+ h(y)

b) differentiate ψ with respect to y to determine h(y)

2. the solution to the DE is ψ(x, y(x)) = c, where c is our constant ofintegration


Additional Example (if time permits)

Determine whether the DE is exact and if it is, solve the DE.

dy

dx=xy2 − cosx sinx

y(1− x2)


Section 2.7 : Substitution Methods




Section 2.7


1. Substitution methods: homogeneous differential equations andBernoulli equations


1) Identify whether a given DE is homogeneous, or Bernoulli

2) Solve homogeneous and Bernoulli equations


Homogeneous

A function f(x, y) is homogeneous function of degree k if

f(λx, λy) = λkf(x, y)


M(x, y) +N(x, y)dy

dx= 0

is homogeneous if and only if M and N are homogeneous functionsof the same order.

Definition: Homogeneous DE

Example: determine whether the differential equation is homogeneous.

(x2 + y2)dx+ (x2 − xy)dy = 0


Solving a Homogeneous DE

If M and N are homogeneous degree k then

M(λx, λy) = λkM(x, y), N(λx, λy) = λkN(x, y),

We can also write this as:


Example 1

Solve the differential equation.

(x2 + y2)dx+ (x2 − xy)dy = 0


Bernoulli Equations


dy

dx+ q(x)y = r(x)yn, n ∈ R

is a Bernoulli equation.

Definition: Bernoulli DE

Observe that

� If n = 1, the DE is separable, linear, and homogeneous.

� The substitution u = y1−n reduces a Bernoulli equation to a linearequation.


Example 2

Solve the DE.

xdy

dx+ y = x2y2


Summary

type form of DE solution

homogeneous M(x, y) + N(x, y) dydx = 0M , N homog. same order

let y = ux or letx = vy

Bernoulli dydx + q(x)y = r(x)yn let u = y1−n


section 2.1 : separable equationspeople.math.gatech.edu/~jbeardsley6/chapter2slides.pdfsection 2.1 :...

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