Only One Word for Review

ReviewEngineering Differential Equations

The Second Test

Euler the Master of Us All

Euler’s Method: Tangent Line Approximation

• For the initial value problem

we begin by approximating solution y = (t) at initial point t0.

• The solution passes through initial point (t0, y0) with slope

f (t0, y0). The line tangent to solution at initial point is thus

• The tangent line is a good approximation to solution curve on an interval short enough.

• Thus if t1 is close enough to t0,

we can approximate (t1) by

0000 , ttytfyy

,)(),,( 00 ytyytfy

010001 , ttytfyy

Euler’s Formula

• For a point t2 close to t1, we approximate (t2) using the line passing through (t1, y1) with slope f (t1, y1):

• Thus we create a sequence yn of approximations to (tn):

where fn = f (tn, yn).

• For a uniform step size h = tn – tn-1, Euler’s formula becomes

nnnnn ttfyy

ttfyy

ttfyy

11

12112

01001

121112 , ttytfyy

,2,1,0,1 nhfyy nnn

Euler Approximation

• To graph an Euler approximation, we plot the points (t0, y0), (t1, y1),…, (tn, yn), and then connect these points with line segments.

Note (t0, y0) was the IC

nnnnnnnn ytffttfyy , where,11

Autonomous Equations and Population Dynamics

• In this section we examine equations of the form y' = f (y), called autonomous equations, where the independent variable t does not appear explicitly.

• y’(t) is the gorillaz velocity which depends on gorillaz height y(t). Important heights are the rest points where y’ is zero; when f(y)=0. These are Equilibrium solutions.

• Example (Exponential Growth):

• Solution:

0, rryy

rteyy 0

Autonomous Equations: Equilibrium Solns

• Equilibrium solutions of a general first order autonomous equation y' = f (y) are found by locating roots of f (y) = 0.

• These roots of f (y) are called critical points.• For example, the critical points of the logistic equation

• are y = 0 and y = K. • Thus critical points are constant

functions (equilibrium solutions)in this setting.

yK

yr

dt

dy

1

Autonomous Equations: Equilibrium Solns

• Equilibrium solutions of a general first order autonomous equation y' = f (y) can be found by locating roots of f (y) = 0.

• These roots of f (y) are called critical points.• Phase diagram is the y axis showing where the monkey

climbs (f>0), rests (f=0) and falls (f<0) y’ = y(10 – y)

• Thus critical points are constant functions (equilibrium solutions)in this setting.

Population Models

• P(t)= fish pop size, b(t) = individ birth rate (births/unit time/fish) d(t) = individ death rate( deaths/unit time/fish)

• Units for P’(t) are fish/unit time• Balance Law

dpdt

ratein rateout bp dp

Velocity and Acceleration

• x(t) height of an object falling in the atmosphere near sea level; time t, velocity v(t) = x’(t), a(t) = x’’(t) accel.

• Newton’s 2nd Law: Net F = ma = m(dv/dt) net force • Force of gravity: - mg downward force• Force of air resistance: - v (opp to v) upward

force• Then get eqn for v (F = Force Grav + Resist Force) and x

mdv

dt mg v

x'v

Velocity and Acceleration

• We can also get one eqn for x (using F = Force Grav + Resist Force)

• m x’’ = -mg – ϒx’ is one second order de for x which is the same as the previous two first order DEs for x and v

mdv

dt mg v

x'v

Homogeneous Equations, Initial Values

• Once a solution to a homogeneous equation is found, then it is possible to solve the corresponding nonhomogeneous equation.

• Thus consider homogeneous linear Diff equations; and in particular, those with constant coefficients (like the previous)

• Initial conditions typically take the form

• Thus solution passes through (t0, y0), and slope of solution at (t0, y0) is equal to y0'.

0 cyybya

0000 , ytyyty

Characteristic Equation

• To solve the 2nd order equation with constant coefficients,

we begin by assuming a solution of the form y = ert. • Substituting this into the differential equation, we obtain

• Simplifying,

and hence

• This last equation is called the characteristic equation of the differential equation.

• We then solve for r by factoring or using quadratic formula.

,0 cyybya

02 rtrtrt cebreear

0)( 2 cbrarert

02 cbrar

General Solution

• Using the quadratic formula on the characteristic equation

we obtain two solutions, r1 and r2. • There are three possible results:

– The roots r1, r2 are real and r1 r2. – The roots r1, r2 are real and r1 = r2. – The roots r1, r2 are complex.

• First assume r1, r2 are real and r1 r2. • In this case, the general solution has the form

,02 cbrar

trtr ececty 2121)(

a

acbbr

2

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Theorem

• Suppose y1 and y2 are solutions to the equation

and that the Wronskian

is not zero at the point t0 where the initial conditions

are assigned. Then there is a choice of constants c1, c2 for which y = c1y1 + c2 y2 is a solution to the differential equation (1) and initial conditions (2).

)1(0)()(][ ytqytpyyL

2121 yyyyW

)2()(,)( 0000 ytyyty

Linear Independence and the Wronskian

• Two functions f and g are linearly dependent if there exist constants c1 and c2, not both zero, such that

for all t in I. Note that this reduces to determining whether f and g are multiples of each other.

• If the only solution to this equation is c1 = c2 = 0, then f and g are linearly independent.

0)()( 21 tgctfc

Theorem

• If f and g are differentiable functions on an open interval I and if W(f, g)(t0) 0 for some point t0 in I, then f and g are linearly independent on I. Moreover, if f and g are linearly dependent on I, then W(f, g)(t) = 0 for all t in I.

Repeated Roots

• Recall our 2nd order linear homogeneous ODE

• where a, b and c are constants. • Assuming an exponential soln leads to characteristic

equation:

• Quadratic formula (or factoring) yields two solutions, r1 & r2:

• When b2 – 4ac = 0, r1 = r2 = -b/2a, since method only gives one solution:

0 cyybya

0)( 2 cbrarety rt

a

acbbr

2

42

atbcety 2/1 )(

General Solution

When roots are the same, then two linearly independent solutions are e^{rt} and te^{rt}

• Thus the general solution for repeated roots is abtabt tececty 2/

22/

1)(

Complex Roots of Characteristic Equation

• Recall our discussion of the equation

where a, b and c are constants. • Assuming an exponential soln leads to characteristic

equation:

• Quadratic formula (or factoring) yields two solutions, r1 & r2:

• If b2 – 4ac < 0, then complex roots: r1 = + i, r2 = - i• Thus

0 cyybya

0)( 2 cbrarety rt

a

acbbr

2

42

titi etyety )(,)( 21

Euler’s Formula; Complex Valued Solutions

• Substituting it into Taylor series for et, we obtain Euler’s formula:

• Generalizing Euler’s formula, we obtain

• Then

• Therefore

titn

ti

n

t

n

ite

n

nn

n

nn

n

nit sincos

!12

)1(

!2

)1(

!

)(

1

121

0

2

0

tite ti sincos

tietetiteeee ttttitti sincossincos

tieteety

tieteetyttti

ttti

sincos)(

sincos)(

2

1

Real Valued Solutions: The Wronskian

• Thus we have the following real-valued functions:

• Checking the Wronskian, we obtain

• Thus y3 and y4 form a fundamental solution set for our ODE, and the general solution can be expressed as

tetytety tt sin)(,cos)( 43

0

cossinsincos

sincos

2

t

tt

tt

e

ttette

teteW

tectecty tt sincos)( 21

Real Valued Solutions

• Our two solutions thus far are complex-valued functions:

• We would prefer to have real-valued solutions, since our differential equation has real coefficients.

• To achieve this, recall that linear combinations of solutions are themselves solutions:

• Ignoring constants, we obtain the two solutions

tietety

tietetytt

tt

sincos)(

sincos)(

2

1

tietyty

tetytyt

t

sin2)()(

cos2)()(

21

21

tetytety tt sin)(,cos)( 43

Theorem (Nonhomogenous Des)

• If Y1, Y2 are solutions of nonhomogeneous equation

then Y1 - Y2 is a solution of the homogeneous equation

• If y1, y2 form a fundamental solution set of homogeneous equation, then there exists constants c1, c2 such that

)()()()( 221121 tyctyctYtY

)()()( tgytqytpy

0)()( ytqytpy

Theorem (General Solution)

• The general solution of nonhomogeneous equation

can be written in the form

where y1, y2 form a fundamental solution set of homogeneous equation, c1, c2 are arbitrary constants and Y is a specific solution to the nonhomogeneous equation.

)()()()( 2211 tYtyctycty

)()()( tgytqytpy

Thanks

• Leonhard Euler for his insights and beautiful mathematics. You are number –e^{(pi)i} in my book.

• Ry Cooder’s music I think it’s going to work (as long as you do)

• Good luck on Test 2

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