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Math 220, Differential Equations

Math 220, Differential Equations. Professor Charles S.C. Lin Office: 528 SEO, Phone: 413-3741 Office Hours: MWF 2:00 p.m. & by appointments E-mail address: cslin@uic.edu. Teaching Assistant. Mr. Diego Dominici Office: 607 SEO Office Hours: ? Phone: 996-4814 e-mail: ddomin1@uic.edu.

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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: 413-3741 Office Hours: MWF 2:00 p.m. & by appointments E-mail address: cslin@uic.edu

  2. Teaching Assistant Mr. Diego Dominici Office: 607 SEO Office Hours: ? Phone: 996-4814 e-mail: ddomin1@uic.edu

  3. Please check: www.math.uic.edu/~berger/M220/index.html for Syllabus, assignments, etc... www.awlonline.com/nagle 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. 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. Direction Fields

  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 graph:

  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. like this: Its direction field

  14. 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. The Method of Isoclines

  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. 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 Application: Mixing Problems (Compartmental Analysis)

  29. 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. 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). Existence and Uniqueness Theorem

  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 62.95 million. Estimate the U.S. population as a function of time.

  34. 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. Application to Newtonian Mechanics

  35. 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. Solution

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

  37. 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. Now, to find the position function x(t)

  38. We obtain: The equation of motion:

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

  40. is the equation Homogeneous equation associated with ()

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

  42. 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. Consider an example

  43. 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. Existence and Uniqueness of 2nd order equation

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

  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.

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