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## diff eq

by: William Houston

16

1

5

# diff eq MECH 2310

William Houston
UTD

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notes
COURSE
Statics
PROF.
P.L. Stephan Thamban
TYPE
Class Notes
PAGES
5
WORDS
CONCEPTS
Calculus
KARMA
25 ?

## Popular in Engineering and Tech

This 5 page Class Notes was uploaded by William Houston on Thursday April 28, 2016. The Class Notes belongs to MECH 2310 at University of Texas at Dallas taught by P.L. Stephan Thamban in Summer 2016. Since its upload, it has received 16 views. For similar materials see Statics in Engineering and Tech at University of Texas at Dallas.

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Date Created: 04/28/16
Lecture Notes for Math 2420 Di▯erential Equations with Applications LECTURE 1 1 What is a Di▯erential Equation? Many of the principles underlying world problems can be stated as relations involving rates at which quantities change. In mathematical terms: Relations ! Equations Rates ! Derivatives of Certain Functions De▯nition 1.1 Equations involving derivatives are called di▯erential equations. Example An object falling in the atmosphere near the sea level. Q: What is a di▯erential equation behind this problem? Let t denote time (measured in seconds) and v (t) = velocity of the falling object at time t (measured in meters/second). Q: What is the physics law governing the motion of the falling object? Using Newton’s Second Law: F = m ▯ a; where F 99K Force acting on the object(newtons) m 99K Mass of the object (kilograms) a 99K Acceleration of the object(meters/second ) As it is very well known dv 0 a = a(t) = dt= v (t) (1) We have the following forces acting on the object 1: Gravity m ▯ g; where g = 9:8 s2 2: Air resistance (drag) ▯ v; where is a constant depending on the falling object Remark: (a) Gravitation acts in the downward direction (positive direction), (b) drag acts in the opposite direction. We have F = m ▯ g ▯ v (2) Combining, 1 and 2 leads to dv m ▯ = m ▯ g ▯ v: (3) dt Equation 3 is called a di▯erential equation of the falling object in the atmosphere near the sea level. Note, that we can write equation 3 as follows dv dt = ▯ m v + g: (4) 1 Specifying ; m one obtains more simple DE (Di▯erential Equation). For example, if m = 4 kg, = 2 kg/sec., then equation 4 takes the form dv 1 = ▯ v + 9:8 (5) dt 2 De▯nition 1.2 Any function v satisfying equation 5 is called a solution to equation 5. Naturally, we may ask the following question: Q: How can we ▯nd solutions to equation 5? Example 2 Mice and owls. Assume a population of mice inhabits some area. We have the following well-known experimental fact: In the absence of predators the population of mice increases at a rate proportional to the current population. Q: What is a di▯erential equation behind this problem? Let t denote time (measured in months) and p(t) = size of population of mice at moment t: Q: What is the law governing the size of the population? dp dt = r ▯ p; (6) where r 99K the proportionality factor (rate constant or growth constant) p 99K size of population of mice at moment t More speci▯cally, for r = 0:5/month dp = 0:5 ▯ p: (7) dt Next, add predators (owls) into the picture: Assume several owls live in the same area and they kill 15 mice per day. Then, equation 7 takes the form dp dt = 0:5 ▯ p ▯ 450 (8) Naturally, we may ask the following question: Q: Why do we have 450 in 8? We observe that since time t is measured in months, while 15 is the information "per day", thus to get the information "per month", we have to multiply 15 by 30: In general, the considered mice-owls model can be de▯ned by the di▯erential equation dp dt = r ▯ p ▯ k (9) where r 99K the proportionality factor (rate constant or growth constant) k 99K constant Q: How can we ▯nd function p(t) satisfying equations 8 and 9? 2 2 Solutions to di▯erential equations 2.1 General solution Consider equation 8: dp(t) 0 = p (t) = 0:5 ▯ p(t) ▯ 450 dt p(t) ▯ 900 p (t) = ; if p(t)6=900; for all t; we have 2 p (t) 1 = : p(t) ▯ 900 2 Using the chain rule d p (t) (lnjp(t) ▯ 900j) = ; hence dt p(t) ▯ 900 Z Z 0 Z d (lnjp(t) ▯ 900j)dt = p (t) dt = 1dt = 1t + C; so dt p(t) ▯ 900 2 2 1 lnjp(t) ▯ 900j = 2 t + C (10) Taking the exponential of the both sides lnjp(t)▯900j 2t+C C 2t e = e = e ▯ e C 2t = e ▯ e ; hence jp(t) ▯ 900j = e ▯ e 2t; and since e ▯ e2t > 0; we have C 2t p(t) ▯ 900 = ▯e ▯ e : Therefore, we have 1t p(t) = 900 + C 1 e 2 ; (11) C where C 1 ▯e is an arbitrary non-zero constant. Recall that, we assumed that p(t) ▯ 900 6= 0 in our approach to solve equation 9. Clearly, the constant function p(t) = 900 for all t; is a solution to this equation, since 1 p (t) = p(t) ▯ 450; so 2 1 (900) 0 = ▯ 900 ▯ 450; thus 2 1 0 = ▯ 900 ▯ 450; and we have 0 = 0: 2 Q: Is the solution p(t) = 900 included in the family of all solutions 1 p(t) = 900 + C ▯1e 2t; C1=6 0: The answer is NO unless we allow for C = 01 Indeed, the substitution for p(t) = 900 into 11, we see that: 1t 900 = 900 + C 1 e 2 ; so 1 t C1▯ e2 = 0 and since e 2t> 0 for all t; we must have C1= 0: 3 Therefore, the solution for p(t) = 900 is included in solutions described by 11 if we allow C 1 = 0. Therefore, 1t p(t) = 900 + C ▯1e 2 , where C 1s any constant describes ALL solutions of equation 8. De▯nition 2.1 A formula describing all solutions to the given DE is called a general solution. Geometric interpretation of general solution Let us consider the family of all functions given by 2t p(t) = 900 + C ▯ 1 ; where C i1 any constant. Figure 1 presents graphs of some functions from this family (integral curves of equation 8). Each graph p 910 905 900 895 890 t -2 -1 1 2 3 4 5 Figure 1: Solutions 11 for C = 11; C = ▯1=1; C = 0; C = 1=2; C = 11 1 in Figure 1 corresponds to one particular choice of the constant C . For in1tance, the choice for the constant C = 0; corresponds to the function 1 p(t) = 900 and the choice for C =11; corresponds to function 2 t p(t) = 900 + e : 2.2 Particular solution: Geometric Interpretation For each value of a constant C in 11, we get a function (curve on the plane) that is called a particular solution or, simply, a solution. Remark: In the considered case di▯erent particular solutions don’t intersect. Furthermore, they can be distinguished by choosing one point such that the graph of the solution passes through it. Example: Find a particular solution of equation 8 passing through the point (1;2): From the general solution, we have p(t) = 900 + C ▯ e 2t; since t = 1 and p(1) = 2; we have 1 1 2 = 900 + C ▯1e 2 ; so ▯ 1 C1 = ▯898e 2 4 Hence, we have 1 1 p(t) = 900 ▯ 898e▯ 2 ▯ et 1 = 900 ▯ 898e 2(t▯1: The problem of ▯nding a particular solution for the given DE (that is a solution passing through a speci▯ed point) is called the Initial Value Problem. In our example above, we can express it as follows: Find a solution for the equation p (t) = 1p(t) ▯ 450 that satis▯es the property p(1) = 2: Therefore, we 2 state the initial value problem for equation 8 as follows: De▯nition 2.2 Consider the following problem ▯ p (t) = 0:5 ▯ p(t) ▯ 450 (12) p(1) = 2 We call it the initial value problem for equation 8. The condition p(1) = 2 is called an initial condition for the initial value problem 12. 5

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