section 1.1 differential equations & mathematical models

Section 1.1 Differential Equations & Mathematical Models

Differential Equations – “Equations with derivatives in them.”

Examples:

1.

2.

3. y″ – 2 = 3x ln(x) + y2

xdy xy edx

243 11

xx yy

What is a solution to a differential equation?

A differential equation will often have infinitely many solutions.

For example, here are some of the many solutions of

1. y = x2

2. y = x2 + e–x

3. y = x2 + 4e–x

4. y = x2 + 31429674e–x

: :

“Family of solutions”

2 2dy y x xdx

Calculus Review:

If then( )dy f xdx

( )y f x dx

A differential equation will usually have infinitely many solutions, but there times when a differential equation will have only one solution or no solutions.

Example: (y ′)2 + y2 = –1

Some differential equations will have solutions, but unfortunately we can't write them down (in terms of our elementary functions).

Example: 2xy e

General Solutions vs. Particular Solutions

1. Solve

2. Solve if y = 5 when x = 1.

3. Solve , y(1) = 5.

23dy xdx

23dy xdx

23dy xdx

Initial Value Problem –

consists of a differential equation along with an initial condition y(xo) = yo

Definition: Order

The order of a differential equation is the order of the highest derivative appearing in it.

Expressing differential equations:

Often we will be able to express 1st order differential equations as ( , ).dy f x ydx

Expressing differential equations:

We will always be able to express. . . .

1st order differential equations in the form F(x, y, y′) = 0

2nd order differential equations in the form F(x, y, y′, y″) = 0 :

nth order differential equations in the form F(x, y, y′, y″, y″′, . . . . , y(n)) = 0

Definition: Solution to a Differential Equation

A function u(x) is a solution to the differential equation F(x, y, y′, y″, . . , y(n)) = 0 on an interval J if u, u′, u″, . . . , u(n) exist on J and F(x, u, u′, u″, . . . , u(n)) = 0 for all x on J.

Ex. 1 (a) Show that y(x) = 1/x is a solution to on the interval [1, 20].13 2xy y

x

Ex. 1 (b) Show that y(x) = 1/x is not a solution to on the interval [-20, 20].13 2xy y

x

Ex. 2 (a) Show that y1(x) = sin(x) is a solution to (y ′ )2 + y2 = 1  

(b) Show that y2(x) = cos(x) is a solution to (y ′ )2 + y2 = 1

Partial Derivatives

Ordinary Differential Equations vs. Partial Differential Equations

Section 1.2 Integrals as General & Particular Solutions

Ex. 1 Solve cosdy xdx

Ex. 2 Solve 21 2 1dyx xdx

Ex. 3 Solve 21 2 1, (1) 3dyx x ydx

Position - Velocity – Acceleration

s(t) = position s′ (t) = velocity s″ (t) = acceleration 

Force = Mass x Acceleration

Ex. 5 A lunar lander is falling freely toward the surface of the moon at a speed of 450 m/s. Its retrorockets, when fired, provide a constant deceleration of 2.5 m/s2 (the gravitational acceleration produced by the moon is assumed to be included in the given deceleration). At what height above the lunar surface should the retro rockets be activated to ensure a "soft touchdown" (velocity = 0 at impact)?

Ex. 5 A lunar lander is falling freely toward the surface of the moon at a speed of 450 m/s. Its retrorockets, when fired, provide a constant deceleration of 2.5 m/s2 (the gravitational acceleration produced by the moon is assumed to be included in the given deceleration). At what height above the lunar surface should the retro rockets be activated to ensure a "soft touchdown" (velocity = 0 at impact)?

Section 1.3 Slope Fields & Solution Curves

Slope field for cosdy xdx

Slope field for 2dy xdx

Ex. 1 Sketch the slope field for y′ = –x

Ex. 2 Sketch the slope field for y′ = x2 + y2

Ex. 3 Examine some solution curves of On the following slope field, draw the solution curve which satisfies the initial condition of. . . . . (a) y(2) = –1  

(b) y(–1) = 3  

(c) y(0) = 0 

(d) y(0) = 1

2 2 0.dyy xdx

Calculus Review (definition of continuity):

f (x) is continuous at xo if

lim ( ) ( )o

ox xf x f x

Calculus Review (definition of continuity):

f (x) is continuous at xo if

f (x, y) is continuous at (xo, yo) if

lim ( ) ( )o

ox xf x f x

, ,lim , ,

o oo ox y x y

f x y f x y

Theorem I: Existence & Uniqueness of SolutionsSuppose that f (x, y) is continuous on some rectangle in the xy-plane containing the point (xo, yo) in its interior and that the partial derivative fy is continuous on that rectangle. Then the initial value problem has a unique solution on some open interval Jo containing the point xo.

( , ); ( ) ,dyo odx f x y y x y

Ex. 4 Determine what this theorem says about the solutions in each of the following differential equations:

(a) 2 53 , (2) 4dy x y ydx

Ex. 4 Determine what this theorem says about the solutions in each of the following differential equations:

(b) 2 2 0, (0) 1dyy x ydx

Ex. 4 Determine what this theorem says about the solutions in each of the following differential equations:

(c) 2 2 0, (1) 0dyy x ydx

Ex. 4 Determine what this theorem says about the solutions in each of the following differential equations:

(d) 1 , (2) 2dy ydx x y

Ex. 4 Determine what this theorem says about the solutions in each of the following differential equations:

(e) 1 , (3) 5dy ydx x y

Ex. 4 Determine what this theorem says about the solutions in each of the following differential equations:

(f) 7 7 , (3) 0dy x y ydx

Section 1.4 Separable Equations & Applications

Definition: Separable Differential Equation

A first order differential equation is said to be separable if

f (x, y) can be written as a product of a function of x and a function of y

(i.e. ). 

( , )dy f x ydx

( ) ( )dy g x h ydx

Definition: Separable Differential Equation

A first order differential equation is said to be separable if

f (x, y) can be written as a product of a function of x and a function of y

(i.e. ). 

Examples:

1.

2.

3.

( , )dy f x ydx

( ) ( )dy g x h ydx

23 1 7 2 3dy x y ydx

2 1sin5

ydy ex xdx y

2

5

3 610

dy x xdx y y

To solve a separable differentiable equation of the form we proceed as follows:

( )( )

dy g xdx h y

To solve a separable differentiable equation of the form we proceed as follows:

h(y) dy = g(x) dx

( )( )

dy g xdx h y

( )( )

dy g xdx h y

To solve a separable differentiable equation of the form we proceed as follows:

h(y) dy = g(x) dx

(Then integrate both sides and solve for y, if this is possible.)

( )( )

dy g xdx h y

( )( )

dy g xdx h y

( ) ( )h y dy g x dx

Ex. 1 Solve 3 24dy x ydx

Ex. 2 Solve dy x ydx

Justification for why this method for solving separable differentiable equations actually works.

Ex. 3 Solve 6 , (0) 7dy xy ydx

Ex. 4 Solvedy xdx y

Review of general solutions and particular solutions.  

Definition: Singular SolutionA particular solution to a first order differential equation is said to be a singular solution if it does not come from the general solution.

Ex. 5 Solve 2/36 1dy x ydx

Ex. 5 Solve 2/36 1dy x ydx

In general:

If we have the differential equation , and h(y) has a zero of yo

then the function y(x) = yo will be a singular solution.

( ) ( )dy g x h ydx

Applications

If y changes at a rate proportional to y then (for some constant k).

Radioactive material & half-lives

dy kydx

Ex. 6 A radioactive substance has a half-life of 5 years. Initially there are 128 grams of this substance. How much remains after t years?

Ex. 7 The half-life of radioactive cobalt is 5.27 years. Suppose that a nuclear accident has left the level of cobalt radiation in a certain region at 100 times the level acceptable for human habitation. How long will it be until the region is again habitable?

Carbon dating

Ex. 8 Carbon extracted from an ancient skull contained only one-sixth as much 14C as carbon extracted from present-day bone. How old is the skull?

Newton's law of cooling (heating)

According to Newton's law of cooling, the time rate of change of the temperature T of a body immersed in a medium of constant temperature A is proportional to

the difference T – A . That is: .dT k T A

dt

Ex. 9 A cake is removed from an oven at 210° F and left to cool at a room temperature, which is 70° F. After 30 min the temperature of the cake is 140° F. When will it be 100° F?

Section 1.5 Linear First-Order Equations

Definition: Linear Differential Equation (First Order)

A first order differential equation is linear if there are functions P(x) and Q(x) so that  

Examples:

1. y′ + sin(x) y = ex

2. y′ – sin(x) y = ex

3. y′ = 3x2y + x3 – 4x + 1

4. cos(x) y′ – sec(x) y = x2

( ) ( )dy P x y Q xdx

Definition: Linear Differential Equation (First Order)

A first order differential equation is linear if there are functions P(x) and Q(x) so that  

Integrating Factor:

( ) ( )dy P x y Q xdx

( )P x dxe

Steps you MUST show when solving a 1st order linear differential equation:

1. Put the differential equation in the form:

2. Compute μ.

3. Multiply μ on both sides of the differential equation to obtain

4. Write this as (μy)′ = μ Q(x)

5. Solve this last differential equation via integration.

( ) ( )dy P x y Q xdx

( ) ( )dy P x y Q xdx

Ex. 1 Solve: dy x ydx

Why isn't there a “+C” in the integrating factor μ?

Ex. 2 Solve: x y′ + 2y = 10x3; y(1) = 5

Reminder: (Theorem I from section 1.3)

Suppose that f (x, y) is continuous on some rectangle in the xy-plane containing the point (xo, yo) in its interior and that the partial derivative fy is continuous on that rectangle. Then the initial value problem

has a unique solution on some open interval Jo containing the point xo.

( , ); ( ) ,o ody f x y y x ydx

Theorem I

If the functions P(x) and Q(x) are continuous on the open interval J containing the point xo, then the initial value problem

has a unique solution y(x) on J .

( ) ( ); ( )o ody P x y Q x y x ydx

Ex. 3 Solve: 2; ( 2) 4yy yx

Reminder of a result from calculus: 1dxdydydx

Applications

Mixture Problems

Solutes, Solvents, & Solutions

Q = amount of solute, V = volume of solution ri = rate in ro = rate out, ci = concentration in, co = concentration out

Ex. 4 Consider a large tank holding 1000 L of water into which a brine solution of salt begins to flow at a constant rate of 6 L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at a rate of 6 L/min. If the concentration of salt in the brine entering the tank is 1 kg/L, determine when the concentration of salt in the tank will reach 0.5 kg/L.

Ex. 5 For the mixture problem described in example 4, assume now that the brine leaves the tank at a rate of 5 L/min instead of 6 L/min and assume that the tank starts out with a concentration of 0.1 kg/L (everything else stays the same as it was in example 4 though). Determine the concentration of salt in the tank as a function of time.

Ex. 6 A swimming pool whose volume is 10,000 gallons contains water that is 0.01% chlorine. Starting at t = 0, city water containing 0.001% chlorine is pumped into the pool at a rate of 5 gal/min, and the pool water flows out at the same rate. What is the percentage of chlorine in the pool after 1 hr? When will the pool be 0.002% chlorine?

Section 1.6 Substitution Methods & Exact Equations

Ex. 1 Solve 29 4dy x ydx

If a differential equation can be written as

let u = Ax + By + C and the resulting differential equation in terms of u and x will be separable.

( ),dy f Ax By Cdx

Definition: First Order Homogenous Differential Equation

A 1st order differential equation that can be expressed as is said to be homogenous. 

Examples:

1.

2.

3.

4.

dy yfdx x

2

3 7dy y ydx x x

2

2

5sin y yyx x

2

2

5sin y xyx y

2 2dyx x ydx

If a differential equation can be written as let

and the resulting differential equation in terms of u and x will be separable.

,dy yfdx x

yux

Ex. 2 Solve 2 2 /y xdyx xy x edx

Here's why a 1st order homogenous differential equation can be turned into a separable differential equation by the substitution .yu

x

Here's why a 1st order homogenous differential equation can be turned into a separable differential equation by the substitution .yu

x

dy yfdx x

Here's why a 1st order homogenous differential equation can be turned into a separable differential equation by the substitution .yu

x

dy yfdx x

( )d ux f udx

Here's why a 1st order homogenous differential equation can be turned into a separable differential equation by the substitution .yu

x

dy yfdx x

( )d ux f udx

( )dux u f udx

Here's why a 1st order homogenous differential equation can be turned into a separable differential equation by the substitution .yu

x

dy yfdx x

( )d ux f udx

( )dux u f udx

( )du f u udx x

Definition: Bernoulli Differential Equation

A 1st order differential equation that can be expressed as is said to be a Bernoulli differential equation. 

Examples:

1.

2.

3.

( ) ( ) ndy P x y Q x ydx

83 sindy xy x ydx

101 345 2

x xy e y yx

2 55tan sinhx y x y x y

Ex. 3 Solve 3 42 (8 5)dy y x ydx x

Ex. 3 Solve 3 42 (8 5)dy y x ydx x

Definition: Exact Differential Equation

A 1st order differential equation that can be expressed as

with is said to be an exact differential equation.

( , ) ( , ) 0dyM x y N x ydx

Exact Differential Equation

M(x, y) dx + N(x, y) dy = 0

( , ) ( , ) 0dyM x y N x ydx

Exact Differential Equation

Solving M(x, y) dx + N(x, y) dy = 0

Ex. 4 Solve (4x – y) dx + (6y – x) dy = 0

Ex. 5 Solve 2 2

2 3

2 3

2 4

xy xdydx x y y

Section 1.8 Acceleration-Velocity Models

Ex. 1 Suppose that an object is a distance of ho from the surface of the earth when it is given an initial velocity of vo. Determine s(t), the position of the object expressed as a function of time. (Assume that the only force acting on the object is due to gravity. Assume a constant acceleration due to gravity of g).

Ex. 2 Suppose that an object is a distance of ho from the surface of the earth when it is given an initial velocity of vo. Determine s(t), the position of the object expressed as a function of time, this time including air resistance. (Use g for the acceleration due to gravity and assume that the force caused by the air resistance is proportional to the velocity.)

Ex. 3 An object of mass 3 kg is released from rest 500 m above the ground and allowed to fall under the influence of gravity. Assume the force due to air resistance is proportional to the velocity of the object with a drag coefficient of 4/3 sec–1. Determine when the object will strike the ground.

Newton's Law of Gravitation –

The gravitational force of attraction between two point masses M and m located at a distance r apart is given by  

2 .GMmFr

Ex. 4 A lunar lander is falling freely toward the surface of the moon at a speed of 450 m/s. Its retrorockets, when fired, provide a constant deceleration of 2.5 m/s2 . At what height above the lunar surface should the retro rockets be activated to ensure a soft touchdown (velocity = 0 at impact)?