Math 220, Differential Equations

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Math 220, Differential Equations

• Professor Charles S.C. Lin
• Office 528 SEO, Phone [phone]
• Office Hours MWF 200 p.m. by appointments
• E-mail address cslin_at_uic.edu

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Teaching Assistant

• Mr. Diego Dominici
• Office 607 SEO
• Office Hours ?
• Phone [phone]
• e-mail ddomin1_at_uic.edu

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Please check

• www.math.uic.edu/berger/M220/index.html for
  Syllabus, assignments, etc...
• www.awlonline.com/nagle for interactive CD

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Differential Equation Classifications

• Ordinary differential equations, order!
• Partial differential equations
• Linear equations i.e. linear in the dependent
  variable(s).
• Nonlinear differential equations not linear
• For example

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Explicit solution and Implicit solution

• If a function satisfies a differential equation,
  for example
• Such a function, defined explicitly as a function
  of independent variable x is called an explicit
  solution. On the other hand, the equation

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Given by

• the following
• The equation
• is said to defined an implicit solution of the
  equation () above.

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In fact, there are many solutions to a D.E such
as () above.

• To find a solution passing through a specific
  point in xy-plane, we need to impose a condition,
  known as initial value, i.e. y(x0) y0. This
  is known as the initial value problem. We shall
  assume that the function f(x,y) is sufficiently
  smooth, that a solution always exists. Namely
• Theorem (Existence and uniqueness) The I.V.P.
• always has a unique solution in a rectangle
  containing the point (x0, y0), if f and f x are
  continuous there.

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Direction Fields

• Consider the first order D. E.
• the equation specifies a slope at each point in
  the xy-plane where f is defined.
• It gives the direction that a solution to the
  equation must have at each point.

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A plot of short line segments drawn at various
points in the xy-plane showing the slope of the
solution curve this is called a direction
field for the differential equation.

• The direction field gives us the flow of
  solutions .

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Example

• For the equation
• Using Maple
• with(DEtools)
• eqdiff(y(t),t)t2-y(t)
• DEplot(eq,y(t),t-5..5,y-5..5,arrowsslim)

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Using Maple program, we have the following

• graph

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Another example

• Consider the logistic equation for the population
  of a certain species
• using maple, we write in commands
• eqdiff(p(t),t)3p(t)-2p(t)2
• DEplot(eq,p(t), t0..5,p0..5,arrowsslim)
• and get

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Its direction field

• like this

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The Method of Isoclines

• Consider the differential equation
• () dy/dx f(x,y).
• The set of points in the xy-plane where all the
  solutions have the same slope dy/dx i.e. the
  level curves for the function f(x,y) are called
  the isoclines for the D. E. ().
• This is the family of curves f(x,y) C.
• This gives us a way to draw direction field.

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Example

• For the differential equation
• f(x,y) x y, and the set of points where
• x y c, are straight lines with slope (-1).
• We can now draw the isoclines for the D. E.
• and the solution passing through a given initial
  point can also be drawn.

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Let us graph the isoclines of f(x,y) x y.

• and compare it to the direction field of it, we
  see

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Maple example

• Let us consider the IVP for y x2 - y, with
  three sets of initial points 0,-1, 0,0 and
  0,2. What will be the corresponding solutions?

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Separable Equation

• Given a differential equation
• If the function f(x,y) can be written as a
  product of two functions g(x) and h(y), i.e.
• f(x,y) g(x) h(y), then the differential eq. is
  called separable.

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Example

• The equation
• is separable, since

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Method for solving separable equation

• Separable equation can be solved easily,
• Rewrite the equation

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Example

• Consider the initial value problem

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Using Maple

• we can solve the IVP with the following Maple
  commands.
• ODEdiff(y(x),x)(y(x)-1)/(x-3)
• ICy(-1)0
• IVPODE,IC
• GSOLNdsolve(ODE,y(x))
• Then use the IC to find the arbitrary constant.

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Linear Equations

• We shall study how one can solve a first order
  linear differential equation of the form
• We first rewrite the above equation in the so
  called standard form

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Integration Factor

• Suppose we multiply a function ?(x) to the above
  equation, we get
• Is it possible for us to find ?(x) such that the
  left hand side
• ?

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• Since
• We see that this can be done, if

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In this case,

• we can solve it by integration.
• Note that

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Examples

• Consider the D.E.
• Solution
• Another example solve the following initial
  value problem

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Application Mixing Problems (Compartmental
Analysis)

• 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

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• Let x(t) be the mass of salt in the tank at time
  t. The rate at which salt enters the tank is
  equal to input rate - output rate. Thus

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The equation is separable

• We can solve it easily, using the initial
  condition, we get

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Existence and Uniqueness Theorem

• Suppose P(x) and Q(x) are continuous on the
  interval (a,b) that contains the point x0. Then
  the initial value problem
• y? P(x)y Q(x), y(x0)y0
• for any given y0.
• has a unique solution on (a,b).

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Application to Population Growth

• If we assume that the growth rate of a population
  is proportional to the population present, then
  it leads to a D.E.
• Let p(t) be the population at time t. Let k gt 0
  be the proportionality constant for the growth
  rate and let p0 be the population at time t
  0. Then a
• mathematical model for a population could be

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This can be solved easily.

• Example In 1790 the population of the United
  States was 3.93 million, and in 1890 it was 62.95
  million. Estimate the U.S. population as a
  function of time.

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Application to Newtonian Mechanics

• The study of motion of objects and the effect of
  forces acting on those objects is called
  Mechanics. A model for Newtonian mechanics is
  based on Newtons laws of motion Let us consider
  an example An object of mass m is given an
  initial velocity of v0 and allowed to fall under
  the influence of gravity. Assuming the
  gravitational force is constant and the force due
  to air resistance is proportional to the velocity
  of the object . Determine the equation of motion
  for this object.

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Solution

• Since the total force acting on the object is
• F FG - FA mg - k v(t). And according to
  Newtons 2nd law of motion, F m a, we see that
  
• m a mg - k v.
• Let x(t) be the position function of the object
  at time t, and
• v(t) dx/dt, a dv/dt.

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Equation of motion can be rewritten as

• The following separable initial value problem.
• We can solve the equation easily, and obtain

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Now, to find the position function x(t)

• Suppose that at t 0, the object is x0 units
  above the ground, i.e. x(0) x0 . Then for the
  position function x(t), we have the following
  I.V.P.
• This can be solved easily.

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We obtain

• The equation of motion

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Linear Differential Operators (4.2)

• We shall now consider linear 2nd order equations
  of the form

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Homogeneous equation associated with (?)

• is the equation

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Remark on linearity of the operator L, and linear
combinations of solutions to homogeneous equation.

• We have L?y1?y2 ?L y1 ?Ly2,
• for any constants ? and ?, and any twice
  differentiable functions y1 and y2 .
• Theroem1. If y1 and y2 are solutions of the
  homogeneous equation
• (HE) y?py?qy0, then any linear
  combination ?y1?y2 of y1 and y2 is also a
  solution of (HE).

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Consider an example

• Ly y? 4y? 3y, We use the convention Dy
  y ?, D2y y ?, Dny y(n) ,
• and rewrite Ly D2y 4Dy 3y, or
  symbolically, L D2 4D 3. Since formerly
  D2 4D 3 (D 3)(D 1), we see that
• Ly (D 3)(D 1)y. The solutions for
• Ly 0 are y1 e-x , and y2 e -3x.

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Existence and Uniqueness of 2nd order equation

• Theorem 2. Let p(x), q(x) and g(x) be continuous
  on an interval (a,b), and x0 ?(a,b). Then the
  I.V.P.

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Fundamental Solutions of Homogeneous Equations

• Let us first define the notion of the Wronskian
  of two differentiable functions y1 and y2. The
  function

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Fundamental solution set

• A pair of solutions y1, y2 of Ly 0, on
  (a,b) where Ly y?py?qy is called a
  fundamental solution set, if Wy1, y2(x0) ? 0
  for some x0 ?(a,b) . A simple example Consider
• Ly y?9y. It is easily checked that y1 cos
  3x and y2 sin 3x are solutions of Ly 0.
  Since the corresponding Wronskian Wy1, y2(x)
  3 ? 0 , thus cos 3x, sin 3x forms a fundamental
  solution set to the homogenenous eq y ? 9y
  0. We see that any linear combination c1 y1 c2
  y2 also satisfies Ly 0. This is known as a
  general solution

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Linear Independence, Fundamental set and Wronskian

• Theorem. Let y1 and y2 be solutions to the
  equation y? py? qy 0 on (a,b). Then the
  following statements are equivalent
• (A) y1, y2 is a fundamental solution set on
  (a,b).
• (B) y1 and y2 are linearly independent on (a,b).
• (C) The Wy1, y2(x) is never zero on (a,b).
• For the proof, we need some linear algebra, i.e.
• Linearly dependent vectors,uniqueness theorem
  etc...

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Reminder

• First Hour Exam
• Date June 15 (Friday)
• Room TBA

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Homogeneous Linear Equations With Constant
Coefficients

• Recall For equations of the form
• ay? by? cy 0,
• by subsituting y e r x, we obtain the
• auxiliary eq ar2 br c 0. If r1 and r2
• are two distinct roots, then a general solution
  is of the form y c1exp(r1x) c2exp(r2x),
  where c1 and c2 are arbitrary constants.

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Repeated Roots

• If in the above equation, r1 r2 r, then a
• general solution is of the form
• y c1exp(rx) c2x exp(rx),
• Example consider the D.E. y? 4y 4 0.
• Its auxiliary equation is r2 4r 4 0,
  hence
• r -2 is a double root, the general solution
  is
• y c1e -2x c2x e -2x,

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Cauchy-Euler Equations

• If an equation is of the form
  ax2y? bxy cy h(x), a, b, c are
  constants, then by letting x e t, we
  transform the original equation into(with t as
  the independent variable), ay ? (b-a)y cy
  h(e t). An equation with constant coefficients.
  Hence can be solved by the method of constant
  coefficients. The equation above is known as a
  Cauchy-Euler Equation.

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Reduction of order

• We know ,in general, a second order linear
  differential equation has two linearly
  independent solutions. If we already have one
  solution, how can we find the other one?
• Let f(x) be a solution to y ? p(x)y q(x)y
  0.
• We will try to find another solution of the form
• y(x) v(x)f(x), with v(x) a non-constant
  function.
• Formerly, we have y vf vf , and y ?
• set w v , etc, we obtain a separable eq. in
  w.
• Finally find v from w by integration.

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Example

• Given f(x) x-1 is a solution to
• x2 y ? - 2xy -4y 0, x gt 0
• find a second linearly independent solution.
• First write the D.E. in standard form.
• Next compute v.
• Finally, 2nd independent solution is y v f.

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Auxiliary Eq. With Complex Roots

• If the auxiliary equation of a linear 2nd order
  D. E. with constant coefficents ar2 br c
  0, has complex roots, (when b2 - 4ac lt 0 ). i.e.
• r1 ? i? and r2 ? - i?, where ? and ?
  are real numbers, then the solutions are
• y1 e (? i?)x , and y2 e (? - i?)x. Since
  we know that e i? x cos ? x i sin ? x , we
  simply take
• y1 e ?x cos ? x , and y2 e ?x sin ? x as
  the two linearly independent solutions.

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And the general solution is of the form

• y(x) c1 e ?x cos ? x c2 e ?x sin ? x, where
  c1 and c2 are arbitrary constants.
• Remark about complex solution
• z(x) u(x) iv(x) to Lz 0 and the fact that
  in this case, we also have Lu 0 and Lv 0.
  Thus the real part and the imaginary part of a
  complex solution to Ly 0 are also solutions
  of Ly 0.

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Example

• Consider the D.E.

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Nonhomogeneous Equation And the the method of
Superposition

• Let L be a linear operator of 2nd order, i.e. L
  D2 pD q, and g ? 0. The equation
• Ly g, is called a Nonhomogeneous eq.
• We wish to solve the equation Ly g , using a
  particular solution to Ly g , and a
  fundamental solution set to Ly 0. First let
  me introduce the concept of the method of
  superposition.

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Theorem Let y1 be a solution to the equation
Ly g1, and let y2 be a solution to the
equation Ly g2, where g1 and g2 are two
functions. Then for any two constants c1 and c2,
the linear combination c1 y1 c2 y2 is a
solution to the equationLy c1 g1 c2 g2 .
(This is known as the Superposition principle).
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Proof
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Representation Theorem of Ly g.

• Theorem Let yp(x) be a particular solution to
  the nonhomogeneous equation () Ly g(x),
  where Ly y ? p(x) y ? q(x) y , on the
  interval (a,b) and let y1(x) and y2(x) be a
  fundamental solution set of Ly 0 on the
  interval (a,b). Then every solution of () can be
  written in the form
• () y(x) yp(x) c1 y1(x) c2 y2(x) .
  This is known as the general solution to ().

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Example

• Given that yp(x) x2 is a particular solution of
  the equation
• () y ? - y 2 - x2,
• find a general solution of ().
• Note the auxiliary equation is r 2 - 1 0. It
  follows that a general solution of () is of the
  form y x2 c1e-x c2e x.

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Superposition Principle the Method of
Undetermined coefficients.

• Example Find a general solution to the D.E.
• Step 1 We first consider the associated
  homogenous equation

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Step 2 Find particular solution to the
Non-homogenous equation using the Superposition
Principle

• There are 2 equations

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To find a particular solution to each of the
above equations

• We use the method of undetermined coefficients,
  that is for the first equation, we try yp
  ax2 bx c, and
• For the second equation, we try yp Aex.
• If any term in the trial expression for yp is a
  solution to the corresponding homogeneous
  equation, then we replace yp by x yp, etc. See
  table 4.1 on Page 208 of your book.

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Next we present a more general method, known as

• The method of variation of parameters,

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The Method of Variation of Parameters

• Consider the non-homogenous linear second order
  differential equation

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Where v1 and v2 are functions to be determined.
We obtain

• two equations (by avoiding 2nd order derivatives
  for the unknows and from Lypg)

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Where v1 and v2 are functions to be determined.
We obtain

• two equations (by avoiding 2nd order derivatives
  for the unknows and from Lypg)

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Finally, solution is found by integration.

• Example

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