Chapter1. Ordinary Differential Equations

Transcription

1 Chapter1. Ordinary Differential Equations Physicists have a variety of reasons for studying differential equations: almost all the elementary and numerous of the advanced parts of theoretical physics are posed mathematically in terms of differential equations. A differential equation is an equation for a function that relates the values of the function to the values of its derivatives. Many real life problems can be formulated as differential equations. For example d 2 y dy + 2x + y = sin(x) dx2 dx is a differential equation. Sometimes one uses shortened notations to write the same equation as or using the differential operator D = dy y + 2xy + y = sin(x) dx D 2 y + 2xDy + y = sin(x) There are two classes of differential equations: O.D.E. (ordinary differential equations): linear and non-linear; An ordinary differential equation (ode) is a differential equation for a function of a single variable, e.g., x(t), P.D.E. (partial differential equations): A PDE equation is a differential equation for a function of several variables, e.g., (x, y, z, t). An ode contains ordinary derivatives and a pde contains partial derivatives. Typically, pde s are much harder to solve than ode s. For example, are ordinary differential equations. y + 4y + y = 0 y = x 2 y + sin(x) u xx + u yy = 0 u 2 x + u 2 y = ln(u) are partial differential equations. (Partial Differential Equations are usually much more difficult). 1. Some concepts related to differential equations: Order: The order of a differential equation is the order of the highest derivative that occurs in the equation. A first order differential equation contains y 0, y and x so it is of the form F(x; y; y 0 ) = 0 or y 0 = f(x; y). For example, the following differential equations are first order: While these are second order: y = x 2 y + e x xy = (1 + y 2 ) y 2 = 4xy + cos(x) y x 2 y + y = 1 + sin(x)e x y + 6yy = (1 + x 3 ) 1

2 General and Particular Solutions: A general solution of a differential equation involves arbitrary constants. In a particular solution, these constants are determined using initial values. As an example, consider the differential equation y = 2x. y = x 2 + c is a general solution, y = x is a particular solution, Example 1. Find the general solution of the differential equation y = 0. Then, find the particular solution that satisfies y(0) = 5; y (0) = 3. y = 0 y = c y = cx + d. Tis is te general solution. y 0 = 3 c = 3, y 0 = 5 d = 5 tus y = 3x + 5 Therefore y = 3x + 5. This is the particular solution. Explicit and Implicit Solutions: y = f(x) is an explicit solution, F(x; y) = 0 is an implicit solution. We have to solve equations to obtain y for a given x in implicit solutions, whereas it is straight forward for explicit solutions. For example, y = e 4x is an explicit solution of the equation y = 4y. x 3 + y 3 = 1 is an implicit solution of the equation y 2 y + x 2 = 0. It is much easier to verify that a function (given either implicitly or explicitly) is a solution of a given differential equation. What is a solution? Solution is a function that satisfied the equation and the derivatives exist. Example 2: Verify that y(t) = e at is a solution of the IVP (initial value problem) y = ay, y(0) = 1. Here y(0) = 1 is called the initial condition. Let s check if y(t) satisfies the equation and the initial condition: They are both OK. So it is a solution. y = ae at = ay, y(0) = e 0 = 1. Example 3: Verify that y(t) = 10 c e t with c a constant, is a solution to y + y = 10. y = ce t = ce t, y + y = ce t + 10 ce t = 10, tus it is verified. Example 4: Verify that y = c 1 sin 2x + c 2 cos(2x) is a solution of y + 4y = 0. Since the second derivative is involved, we differentiate y twice to get and substituting these in the given equation, 2

3 Example 5: Verify that the function x 2 = 2 y 2 ln(y) solves the equation y = We differentiate both sides of the given solution to get xy (x 2 +y 2 ). Linear or non-linear equations: Let y(t) be the unknown. Then, a 0 t y n + a 1 t y n a n t y = g t (1) is a linear equations. If the equation can not be written as Eq. 1, then it s a non-linear. Two things you must know: identify the linearity and order of an equation. Example 6: Let y(t) be the unknown. Identify the order and linearity of the following equations. 2. First-order differential Equations The general first-order differential equation for the function y = (x) is written as dy dx = f(x, y) (2) where f(x, y) can be any function of the independent variable x and the dependent variable y. We first show how to determine a numerical solution of this equation, and then learn techniques for solving analytically some special forms of (2), namely, separable, exact and linear first-order equations. a) Separable equations: A differential equation is said to be separable if it can be rewritten so that terms involving the differential of y is on one side of the equation, and those of x on the other side. One then integrates to get rid of the differentials leading to an equation that implicity or explicitly gives y. Thus, g(y)dy = f(x)dx, it is called a separable equation. We can find the solution by integrating both sides. (Don't forget the integration constant!). 3

4 g y dy = f x dx + c Example 1: Solve the initial value problem y = x 3 e y, with y(1) = 0. Example 2: Solve y 2 y + 2x = 0. Example 3: Find the particular solution of the equation y ln y dx xdy = 0 such that when x = 1, y = 2. Rewrite the equation as 4

5 b) Exact equations: A first order equation can often be written in the form M (x; y) dx + N (x; y) dy = 0 This equation is said to be exact if the left hand side is the differential of a function u(x; y) so that u(x, y) x = M x, y, u(x, y) y = N x, y, In other words, du = Mdx + Ndy, so Mdx + Ndy is a total differential. For example, the equation is exact and (4x 3 + 2xy 2 )dx + (4y 3 + 2yx 2 )dy = 0 u x, y = x 4 + x 2 y 2 + y 4 So, the solution of this equation is very simple, if du is zero, u must be a constant, therefore c = x 4 + x 2 y 2 + y 4 Hence, from the equations related with M and N, the condition M y = N x is necessary and sufficient for the equation M(x; y)dx + N(x; y)dy = 0 to be exact. Example 1: Solve the equation 3y 2 dx + (3y 2 + 6xy)dy = 0 5

6 Example 2: Solve 2x sin(y) dy + x 2 cos(y) dy = 0. And Integrating Factors Consider the equation P dx + Q dy = 0 that is not exact. If it becomes exact after multiplying by, i.e. if (x)pdx + (x)qdy = 0 is exact, then is called an integrating factor. (Note that P, Q and are functions of x and y). Example: Solve (2xe x 3y 2 )dx + 2ydy = 0. Use integrating factor as e x. 6

7 Now the equation is exact. We can solve it as we did the previous example and obtain the result How to Find Integrating Factors? x 2 + y 2 e x = c The first-order linear differential equation (linear in y and its derivative) can be written in the form y + p(x)y = g(x), with the initial condition y(x 0 ) = y 0. Linear first-order equations can be integrated using an integrating factor (x). After mathematical calculation, the solution of above equation) satisfying the initial condition (x 0 ) = y 0 is then traditionally written as Example 1: Solve dy dx + 2y = e x with y(0) = 3/4. Note that this equation is not seperable. With p(x) = 2 and g(x) = e x, we have 7

8 Example 2: Solve y + ay = b. Example 3: Solve ty + 2y = 4t 2 with initial condition y(1) = 2. c) Bernoulli Equation: The equation y + p x y = g(x)y α is called Bernoulli equation. It is nonlinear. Nonlinear equations are usually much more difficult than linear ones, but Bernoulli equation is an exception. It can be linearized by the substitution u x = [y(x)] 1 α Then, we can solve it as other linear equations. Example: Solve the equation 8

9 3. Mathematical Modeling Differential equations are the natural tools to formulate, solve and understand many engineering and scientific systems. The mathematical models of most of the simple systems are differential equations. The simplest ordinary differential equations can be integrated directly by finding antiderivatives. Example 1: Malthus' law of population Dynamics; We all know that population grows in general, but how fast does the population of a city grow? What will the population of the US be in 2025? Questions of this kind have been of interest to many scientists and the first person to propose a mathematical law was Rev. Thomas Robert Malthus, an English clergy man who laid out his findings in his 1798 writings. Malthusian law says: The rate of change of population is proportional to the actual population at any given time. Notice that this law is very similar in nature to Newton's law. Both laws predict that the rate of change of a quantity is in some sense proportional to the quantity remaining at a given time. The only difference is that Newton's law has to do with the decrease in heat while Malthus law has to do with increase in population. Let the population at a given time t 0 be P 0 and at any time t later be P(t). Then according to Malthus law, dp dt P If k is the constant of proportionality, the above equation becomes dp dt = kp This is a very simple equation whose solution is ln P = k t + c, where c is the constant of integration. Using the initial condition that at t = t 0 ; P = P 0, we get We can then write the solution as that is, ln P 0 = k t 0 + c ln P = k t + ln P 0 k t 0 ln P = k(t t 0 ) + ln P 0 P = P 0 k(t t 0 ) ** The world population in 1965 and in 1970 was billions and billions, respectively. What was the population in 1973? Solution: From equation (2.19), using the data for 1965 and 1970, 9

10 To find the population in 1973, we have that The actual population of the world in 1973 was billions! Example 2: Newton s Law of motion. Consider a mass falling under the influence of constant gravity, such as approximately found on the Earth s surface. Newton s law, F = ma, results in the equation where x is the height of the object above the ground, m is the mass of the object, and g = 9.8 meter/sec2 is the constant gravitational acceleration. Find the height as a function of time if the initial height is y 0 and initial speed is v 0. Here, the right-hand-side of the ode is a constant. The first integration, obtained by antidifferentiation, yields where A is the constant of integration, and the second integration gives with B another constant of integration. The two constants of integration A and B can be determined from initial conditions. If we know that the initial height of the mass is x0, and the initial velocity is v0, then the correct initial conditions are Substitution of these initial conditions into the equations for dx/dt and x allows us to solve for A and B to determine the unique solution that satisfies both the ode and the initial conditions: 10

11 Example 3: Exponential growth/decay. Assume the rate of change of Q(t) is proportional to the quantity at time t. We can write where r is the rate of growth/decay and Q(t) is the amount of quantity at time t. If r > 0: exponential growth If r < 0: exponential decay Differential equation: Here r is called the growth rate. By IC, we get Q(0) = C = Q 0. The solution is Two concepts: Doubling time TD (only if r > 0): is the time that Q(T D ) = 2Q 0. Half life (or half time) TH (only for r < 0): is the time that Q(T H ) = (½) Q 0. Note here that T H > 0 since r < 0.. NB! TD, TH do not depend on Q 0. They only depend on r. Example: A radio active material is reduced to 1/3 after 10 years. Find its half life. Model: dq/dt = rq, r is rate which is unknown. We have the solution Q(t) = Q 0 e rt. So Q 10 = Q 0 e rt = Q 0, Q 3 0e 10r = Q 0 ln 3, r = 3 10 To find the half life, we only need the rate r T H = ln 2 10 ln 2 = ln 2 = 10 r ln 3 ln 3 11

12 Review Examples

13 4. 13

14 Exercises 1. Show that the equation is exact, and obtain its general solution. Also, find the particular solution corresponding to the given initial condition as well. Ans: 2. Obtain the general solution, using the methods of this section. Ans: a) x 2-2xy y 2 = C 3. Use separation of variables to find the general solution. Ans: 4. Solve y = (3x 2-1)/(2y), subject to the given initial condition a) y(0) = -3, b) y(0) = -1, c) y(4) = 5, d) y(-1)=0. Ans: 5. Solve x y' + y= 6x 2 subject to the given initial condition a) y(1) = 0 and b) y(-3) = 18, using any method of this section. Ans: a) y(x) = 2x 2 2/x, b) y(x) = 2x 2 14

15 6. Solve the problems in the course books. References: 1. Greenberg, M.D. Advanced Engineering Mathematics, 2nd edition. Prentice Hall, O'Neil, P.V. Advanced Engineering Mathematics, 5th edition. Thomson, Ross, S.L. Introduction to Ordinary Differential Equations, 4th edition. Wiley,