Differential Equations Introduction to Differential Equations
Only One Word for Review Review Engineering Differential Equations The Second Test.
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Transcript of Only One Word for Review Review Engineering Differential Equations The Second Test.
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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