Math 20D Final Exam Reviek3walsh/math20d/FinalReview.pdf · 2012. 8. 2. · Title: Math 20D Final Exam Review Author: Katie Walsh Created Date: 8/1/2012 6:07:21 PM

Exam Basics

1 There will be 9 questions.

2 The first 3 are on pre-midterm material.

3 The next 1 is a mix of old and new material.

4 The last 5 questions will be on new material since themidterm.

5 60 points total (+3 bonus points)

Katie Walsh Math 20D Final Exam Review

First Order Differential Equations

1 Linear First Order Differential Equations1 y ′ + p(t)y = g(t)2 Let µ(t) = e

∫p(t)dt

3 y=1/µ(t)∫µ(t)g(t)dt

2 Separable First Order Differential Equations1 M(x) + N(y)dy/dx = 02∫

N(y)dy = −∫

M(x)dx


Exact and Integrating Factor for Exact

A differential equation of the form

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

is exact ifMy = Nx

where My is the partial derivative of the function M(x , y) withrespect to y and Nx is the partial derivative of N(x , y) withrespect to y .If the equation is exact then we can find some function f such thatfx = M and fy = N. Then the solution to the differential equationis f = C .


Sometimes an equation is not exact but we can use an integratingfactor to make it exact. There are two types of problems youshould be able to do.

1 Equations that can be made exact by multiplying by a functionµ(x) that only depends on x .This is true only if

My − Nx

N

only depends on N. In this case

µ(x)′ =My − Nx

Nµ(x)

e.g.y ′ = e2x + y − 1


2 Equations that can be made exact using a given integratingfactor.e.g.

x2y3 + x(1 + y2)y ′ = 0

µ(x , y) = 1/(xy3)


Examples of first order differential equations

1 dy/dx = x3−2yx

2 dy/dx = 2x+y3+3y2−x

3 dy/dx = 4x3+1y(2+3y)

4 For more examples see Pgs 132-133


Autonomous Equations

1 Graph dy/dt as a function of y . Focus on zeros and wherethe function is positive and negative.

2 Graph the phase line (number line with equilibrium pointsmarked and draw arrows to represent when solutions areincreasing or decreasing)

3 Classify the stationary points as stable, unstable orsemi-stable.

4 Sketch several solutions based on the increasing anddecreasing information from the phase line.

Example: dy/dt = y2(y2 − 1) where −∞ < y0 <∞


Figure: Plot of dy/dt as a function of y

0

-1

1

-2

2

stable

semistable

unstable

Figure: Phase Line and sketch of solutionsKatie Walsh Math 20D Final Exam Review

Figure: Actual Solutions

Follow-up Question: What happens to y as t →∞?This depends on the initial condition! Let y(t0) = y0.For y0 > 1, y →∞.For 0 < y0 < 1, y → 0.For y0 < 0, y → −1 .


Theorems about guaranteed solution intervals

First Order

Theorem

If p and q are continuous on an interval I : α < t < β containingthe point t = t0, then there exists a unique function y thatsatisfies the IVP

y ′ + p(t)y = g(t)

y(t0) = y0

for each t in I .

e.g. Find the largest interval in which the IVP

ty ′ + 2y = 4t2, y(1) = 2

has a unique solution.


Theorem

Let f (t, y) and ∂f /∂y be continuous for α < t < β, γ < y < δcontaining the point (t0, y0). Then in some intervalt0 − h < t < t0 + h contained in α < t < β, there is a uniquesolution to the IVP y ′ = f (t, y), where y(t0) = y0

e.g. Does the theorem guarantee that y ′ = y1/3 with the initialcondition y(0) = 0 has a unique solution?No, since ∂f /∂y is undefined at y = 0. In fact, y = (2/3t)3/2 andy = −(2/3t)3/2 are solutions to the IVP.


Theorems for 2nd Order

Theorem

Consider the IVP

y ′′ + p(t)y ′ + q(t)y = g(t), y(t0) = y0, y ′(t0) = y ′0

where p, q, g are continuous on an open interval I containing t0.Then there exists a unique solution y to this problem and thesolution exists throughout I .


Theorems about the Wronskian

Theorem

Suppose that y1 and y2 are two solutions of the differentialequation

y ′′ + p(t)y ′ + q(t)y = 0.

Then the family of solutions y = c1y1(t) + c2y2)t) with arbitrarycoefficients c1 and c2 includes every solutions to the abovedifferential equation if and only if there is a point t0 where theWronskian of y1 and y2 is not zero.

If y1 and y2 are both solutions to a homogeneous second orderdifferential equation and their Wronskian is nonzero for some pointt0, we call y1 and y2 a fundamental solution set.


Just because we can write every solutions as y = c1y1 + c2y2doesn’t mean we can solve every initial value problem. (Why?Because every solution can be expressed via infinitely many initialconditions. Say the solution to the IVP is y(t) = 2t. This satisfiesthe initial conditions y(0) = 0, y(1) = 2, y(7) = 14, . . . . Some ofthese t0 may have W (y1, y2)(t0) = 0

Theorem

Lety ′′ + p(t)y ′ + q(t)y = 0,

have solutions y1 and y2. The the IVP with initial conditionsy(t0) = y0, y

′(t0) = y ′0 has a solutions of the form y = c1y1 + c2y2if and only if W (y1, y2)(t0) 6= 0


Second Order, Linear, Homogeneous Differential Equationswith Constant Coefficients

ay ′′ + by + c = 0

Characteristic Polynomial ar2 + br + c = 0Let r1 and r2 be the roots of the characteristic polynomial.

1 If r1 and r2 are real and distinct, then Let y1 = er1t andy2 = er2t .

2 If r1, r2 = λ+ µi then y1 = eλtcosµt and y1 = eλtsinµt.

3 If r1 = r2 then y1 = er1t and y2 = ter1t .

In each case, y1 and y2 form a fundamental solution set so thegeneral solution is y = c1y1 + c2y2.


Examples

1 y ′′ − 2y ′ + y = 0

2 y ′′ − y ′ − 6y = 0

3 y ′′ − 10y ′ + 26y = 0


Method of Undetermined Coefficients

Given the differential equation

y ′′ + p(t)y ′ + q(t)y = g(t)

we can use the method of undetermined coefficients to find thesolutions. First we find the fundamental solution set to thecorresponding homogeneous equation. Then we guess the form ofthe particular solutions.

1 If g(t) = A0tn + A1tn−1 + · · ·+ An−1t + An, guessB0tn + B1tn−1 + · · ·+ Bn−1t + Bn

2 If g(t) = eαt , guess Aeαt

3 If g(t) = sin(αt) or g(t) = cos(αt), guessAsin(αt) + Bcos(αt).

If we have a product of the above, guess the product (make sure tohave different constants or functions in front of the sin or cosparts.) If we have a sum, break into different problems, finding ag(t) for each term in the sum.


Warning! If the guess is one of the homogeneous solutions,multiply the guess by t. If it is still a solution to the homogeneoussolution (in the case of repeated roots), multiply by t again.


What should we guess for the following:

1 −3te−t

(At + B)e−t

2 t2cos(2t)(At2 + Bt + C )cos(2t) + (Dt2 + Et + F )sin(2t)

3 et(sin(5t))Aet(cos(5t)) + Bet(sin(5t))

Once we have our guess, we plug it back in to the differentialequation. Then we can solve for the coefficients. Example:

y ′′ − y = t2 + 3et


Giveny ′′ + p(t)y ′ + q(t)y = g(t)

we can use variation of parameters to solve. We first find y1 andy2 the fundamental solution set. Then the particular is of the form

Y (t) = −y1

∫y2(t)g(t)

W (y1, y2)dt + y2(t)

∫y1(t)g(t)

W (y1, y2)dt

As long as we remember our “+C ’s” this will in fact be thegeneral solution. Example: Consider

t2y ′′ − 2y = 3t2 − 1

Show that the corresponding homogeneous equation hasfundamental solution set y1(t) = t2, y2(t) = t−1.Use variation of parameters to find the general solution.


New stuff since the first midterm

1 Systems of Linear Differential Equations

2 Series Solutions

3 Laplace Transforms


Systems of Linear Differential Equations

1 Know how to find eigenvalues and eigenvectors of a 2x2matrix.

Example: Find eigenvalues and eigenvectors of A =

[2 −51 −2

]2 Know how to find the general solution set to x′ = Ax in each

of the three cases below.Case 1:If Anxn has n-linearly independent eigenvectors.

e.g. A has λ1 = −1 with v1 =

(12

)and λ2 = −4 with

v2 =

(1−4

). What is the general solution?

x = c1

(12

)e−t + c2

(1−4

)e−4t .


Case 2If A2x2 has complex conjugates as its eigenvalues.

Example: A has λ = 1± 2i with eigenvector v =

(1± i−4

)Let x1 =

(1 + i−4

)e1+2i . Expand into real and imaginary parts.

Then the general solution is

x = c1(real part ofx1) + c2imaginary part ofx1


Case 3If A2x2 has a repeated eigenvalue with only one linearlyindependent set of eigenvectors, we need to find the generalizedeigenvector.

Example:Let A =

[1 −11 3

]. Then A has eigenvalue 2 with(

1−1

)as its only linearly independent eigenvector. Then we

have to find the generalized eigenvector by solving (A− λI )w = vi.e.

(A− 2I )w =

(1−1

)The vector w will always have the form w = av + g where a is anarbitrary constant. Then the general solution is

x = c1veλt + c2(vteλt + geλt)


Theorems for Systems of First Order Linear Equations

Theorem

If x(1), x(2), · · · , x(n) are solutions to the system x′ = P(t)x onα < t < β and W [x(1), x(2), · · · , x(n)] 6= 0 for some point in thatinterval, then x(1), x(2), · · · , x(n) form a fundamental set ofsolutions for that interval.


Series Solution

Given a second order linear homogeneous differential equation withpolynomial terms P(x)y ′′ + Q(x)y ′ + R(x)y = 0 we can use seriessolution.Assume that the solution has the form

y =∞∑n=0

an(x − x0)n

We need that x0 is an ordinary point of the differential equation,i.e. P(x0) 6= 0. Then

y ′ =∞∑n=0

nanxn−1

y ′′ =∞∑n=0

n(n − 1)anxn−2


Plug this in to the differential equation. Then we re-index asnecessary and add. We can then find a recurrence relation for thecoefficients. The first two coefficients are free, say a0 and a1. If welook at the terms with a0 we get one solution, the other solutionare the terms with a1.Example: (1− x)y ′′ + y = 0 with x0 = 0Find the recurrence relation and the first four terms of each of thetwo solutions.


Theorem

If x0 is an ordinary point of P(x)y ′′ + Q(x)y ′ + R(x)y = 0 andQ(x)/P(x) and R(x)/P(x) have a Taylor Series at x0 then we geta general solution

y =∞∑n=0

an(x − x0)n = a0y1(x) + a1y2(x)

where a0 and a1 are arbitrary. The radius of convergence for y1and y2 is at least as large as that of Q(x)/P(x) and R(x)/P(x).

Example: Find lower bound on radius of convergence of(1 + x3)y ′′ + 4xy ′ + y = 0 about x0 = 0 and x0 = 2.


Euler Equations

Euler Equations have the form

ay ′′ + by ′ + cy = 0

These have x0 = 0 as a singular point. We guess that the solutiony = x r . Plugging this in we get

x r (ar2 + (b − a)r + c) = 0

We then find the roots of the quadratic equations.


If the roots are real and distinct r = r1, r2

y = c1|x |r1 + c2|x |r2

If the roots are repeated r = r1

y = c1|x |r1 + c2|x |r1 ln(x)

If the roots are complex r = λ± µi

y = c1|x |λ cos(µ ln |x |) + c2|x |λ sin(µ ln |x |)


Laplace Transform

Theorem

If f is a piecewise continuous on o ≤ t ≤ A for all A > 0 and|f (t)| ≤ Ceat for t ≥ M then the Laplace transform of f is

L{f (t)} =

∫ ∞0

e−st f (t)dt

The Laplace transform is linear.

L{c1f1 + c2f2} = c1L{f1}+ c2L{f2}


L{f ′(t)} = sL{f (t)} − f (0)

L{f ′′(t)} = s2L{f (t)} − sf (0)− f ′(0)

L{f (n)(t)} = snL{f (t)} − sn−1f (0)− · · · − sf (n−2)(0)− f (n−1)(0)

We can use these facts to solve differential equations. First takeLaplace transform of both sides using linearity. Then solve forL{f (t)}. Then use table to find inverse Laplace Transform. Youmay need to use partial fractions and complete the square.Example: y ′′ − 2y ′ + 2y = cos t with y(0) = 1 and y ′(0) = 0


The Unit Step Function (Heaviside Function)

For c > 0, we can define the unit stpe function at c .

uc(t) =

{0 : t < c1 : t ≥ c

Then

L{uc(t)} =e−cs

s

andL{uc(t)f (t − c)} = e−csL{f }

We can use this to solve differential equations with discontinuosforcing functions. E.g. y ′′ + 3y ′ + 2y = u2(t)


The Dirac Delta Function

The dirac δ function is defined to satisfy the following properties:

δ(t − t0) = 0 t 6= t0∫ ∞−∞

δ(t − t0) = 1

We can use Laplace Transforms to solve differential equationsinvolving δ(t − t0)

L{δ(t − t0)} = e−st0

Example: y ′′ + 2y ′ + 2y = δ(t − π)


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Math 20D Final Exam Reviek3walsh/math20d/FinalReview.pdf · 2012. 8. 2. · Title: Math 20D Final Exam Review Author: Katie Walsh Created Date: 8/1/2012 6:07:21 PM