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Math 220, Differential Equations

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1 Math 220, Differential Equations
Professor Charles S.C. Lin Office: 528 SEO, Phone: Office Hours: MWF 2:00 p.m. & by appointments address:

2 Teaching Assistant Mr. Diego Dominici Office: 607 SEO Office Hours: ?

3 Please check: for Syllabus, assignments, etc... for interactive CD

4 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:

5 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

6 Given by the following: The equation
is said to defined an implicit solution of the equation (*) above.

7 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.

8 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.

9 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 ”.

10 Example For the equation Using Maple with(DEtools);
eq:=diff(y(t),t)=t^2-y(t); DEplot(eq,y(t),t=-5..5,y=-5..5,arrows=slim)

11 Using Maple program, we have the following

12 Another example: Consider the logistic equation for the population of a certain species: using maple, we write in commands: eq:=diff(p(t),t)=3*p(t)-2*p(t)^2; DEplot(eq,p(t), t=0..5,p=0..5,arrows=slim); and get

13 Its direction field like this:

14 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.

15 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.

16 Let us graph the isoclines of f(x,y) = x + y.
and compare it to the direction field of it, we see

17 Maple example Let us consider the IVP for y’ = x^2 - y, with three sets of initial points: [0,-1], [0,0] and [0,2]. What will be the corresponding solutions?

18 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.

19 Example The equation is separable, since

20 Method for solving separable equation
Separable equation can be solved easily, Rewrite the equation:

21 Example Consider the initial value problem

22 Using Maple: we can solve the IVP with the following Maple commands.
ODE:=diff(y(x),x)=(y(x)-1)/(x-3); IC:=y(-1)=0; IVP:={ODE,IC}; GSOLN:=dsolve(ODE,y(x)); Then use the IC to find the arbitrary constant.

23 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”:

24 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 ?

25 Since We see that this can be done, if

26 In this case, we can solve it by integration. Note that:

27 Examples: Consider the D.E. Solution:
Another example: solve the following initial value problem:

28 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

29 Let x(t) be the mass of salt in the tank at time t
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

30 The equation is separable
We can solve it easily, using the initial condition, we get

31 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).

32 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 > 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:

33 This can be solved easily.
Example: In 1790 the population of the United States was 3.93 million, and in 1890 it was million. Estimate the U.S. population as a function of time.

34 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 Newton’s 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.

35 Solution Since the total force acting on the object is
F = FG - FA = mg - k v(t). And according to Newton’s 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.

36 Equation of motion can be rewritten as:
The following separable initial value problem. We can solve the equation easily, and obtain:

37 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.

38 We obtain: The equation of motion:

39 Linear Differential Operators (4.2)
We shall now consider linear 2nd order equations of the form:

40 Homogeneous equation associated with ()
is the equation

41 Remark on linearity of the operator L, and linear combinations of solutions to homogeneous equation.
We have L[y1+y2]= L[ y1]+ L[y2], 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+qy=0, then any linear combination y1+y2 of y1 and y2 is also a solution of (HE).

42 Consider an example L[y] = y + 4y + 3y, We use the convention Dy = y , D2y = y , Dny = y(n) , and rewrite L[y] = D2y + 4Dy +3y, or symbolically, L = D2 + 4D +3. Since formerly D2 + 4D +3 = (D + 3)(D + 1), we see that: L[y] = (D + 3)(D + 1)[y]. The solutions for L[y] = 0 are y1 = e-x , and y2 = e -3x.

43 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.

44 Fundamental Solutions of Homogeneous Equations
Let us first define the notion of the Wronskian of two differentiable functions y1 and y2. The function

45 Fundamental solution set
A pair of solutions [y1, y2] of L[y] = 0, on (a,b) where L[y] = y+py+qy is called a fundamental solution set, if W[y1, y2](x0)  0 for some x0 (a,b) . A simple example: Consider L[y] = y+9y. It is easily checked that y1 = cos 3x and y2 = sin 3x are solutions of L[y] = 0. Since the corresponding Wronskian W[y1, 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 L[y] = 0. This is known as a general solution

46 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 W[y1, y2](x) is never zero on (a,b). For the proof, we need some linear algebra, i.e. Linearly dependent vectors,uniqueness theorem etc...

47 Reminder First Hour Exam: Date: June 15 (Friday) Room: TBA

48 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.

49 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,

50 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.

51 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 ´ = v´f + vf ´, and y  = … set w = v ´, etc…, we obtain a separable eq. in w. Finally find v from w by integration.

52 Example Given f(x) = x-1 is a solution to
x2 y  - 2xy´ -4y = 0, x > 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.

53 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 < 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.

54 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 L[z] = 0 and the fact that in this case, we also have L[u] = 0 and L[v] = 0. Thus the real part and the imaginary part of a complex solution to L[y] = 0 are also solutions of L[y] = 0.

55 Example Consider the D.E.

56 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: L[y] = g, is called a Nonhomogeneous eq. We wish to solve the equation L[y] = g , using a particular solution to L[y] = g , and a fundamental solution set to L[y] = 0. First let me introduce the concept of the method of superposition.

57 Theorem: Let y1 be a solution to the equation L[y] = g1, and let y2 be a solution to the equation L[y] = 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 equation L[y] = c1 g1 + c2 g2 . (This is known as the Superposition principle).

58 Proof

59 Representation Theorem of L[y] = g.
Theorem: Let yp(x) be a particular solution to the nonhomogeneous equation (*) L[y] = g(x), where L[y]= 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 L[y] = 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 (*).

60 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 = 0. It follows that a general solution of (*) is of the form y = x2 + c1e-x + c2e x.

61 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:

62 Step 2: Find particular solution to the Non-homogenous equation using the Superposition Principle
There are “2” equations:

63 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.

64 Next we present a more general method, known as:
The method of variation of parameters,

65 The Method of Variation of Parameters
Consider the non-homogenous linear second order differential equation :

66 Where v1 and v2 are functions to be determined. We obtain:
two equations (by avoiding 2nd order derivatives for the unknows and from L[yp]=g):

67 Where v1 and v2 are functions to be determined. We obtain:
two equations (by avoiding 2nd order derivatives for the unknows and from L[yp]=g):

68 Finally, solution is found by integration.


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