Exam Question 10: Differential Equations. June 19, Applied Mathematics: Lecture 6. Brendan Williamson. Introduction.

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1 Exam Question 10: June 19, 2016

2 In this lecture we will study differential equations, which pertains to Q. 10 of the Higher Level paper. It s arguably more theoretical than other topics on the syllabus, and relies heavily on integration. Because of this I will split up this lecture into three parts; first I will give a brief introduction to integration, in case any of you are studying this before you learned integration in maths (if not you should skip it, as I won t be doing anything special and in fact won t go into as much detail as your maths class). We will then move on to differential equations and how to solve them; this will again be an exercise in pure theory. Finally we apply this knowledge to real world problems involving force, motion and other topics.

3 : Motivation It is often the case that we are interested in a function but are instead given its derivative. For example, you could be told that the acceleration of an object at time t is equal to a(t) = 5 2t, but you want to know the velocity v. So we re asking What function has derivative 5 2t? One answer is 5t t 2. Another is 5t t In fact 5t t 2 + C has derivative 5 2t for all values of C, so v could be any of these functions. We will call such functions involving C general anti-derivatives. In general, if F and f are two functions and F (t) = f (t) for all t, we say that F is an anti-derivative of f. The family of anti-derivatives of f is denoted by f (t). This notation may seem arbitrary, and to a certain extent it is, but one of the main motivations for using anti-derivatives (albeit one that we won t use here) is for finding areas under curves and it makes more sense if you study integration in that context. Because we won t do that I ll instead offer this rough motivation for the term.

4 : Motivation First ask yourself, what is the general anti-derivative for 1? If we re talking about functions of t, then it s t + C. If we re talking about functions of x then it s x + C. So if you want the anti-derivative for f (t) = at where a is come constant, then when you write at the term is a reminder that t is the variable in the function and a is just some constant. So if you see 1 dx = dx, you know the answer is x + C. This answers the question of why you need some reference to t in the notation for the anti-derivative, but why specifically? As you may suspect, it is related to the notation f (t) = df that we use for derivatives. We know that f (t) = f (t) + C. However doing some of the calculations you may have done when you learned the chain rule for derivatives, we could argue that df f (t) = = df = f + C.

5 of Elementary Functions It will of course be useful to know the anti-derivatives to some basic functions. Below is a table of the anti-derivatives of many of the elementary functions; you will have a similar list in your mathematical tables. f (x) f (x) dx x n, n 1 x n+1 n+1 + C 1 a+bx e ax 1 b ln a + bx + C 1 a eax + C b ax 1 a ln(b) bax + C sin(ax) 1 a cos(ax) + C cos(ax) 1 a sin(ax) + C 1 a 2 +x 2 1 a tan 1 ( x a ) + C 1 a 2 x 2 sin 1 ( x a ) + C

6 Anti-Differentiation Techniques There is no great mathematical theory that goes in to finding these anti-derivatives; once you know that the derivative of x 2 is 2x, you should be able to figure out that the anti-derivative of x is x2 2 + C just by guessing and checking. In fact all of the above can be verified by guessing and checking. Just like for differentiating, there is one useful formula to remember when finding anti-derivatives. It is easy to show that if F is an anti-derivative of f then 2F is an anti-derivative of 2f, and if G is an anti-derivative of g then F + G is an anti-derivative of f + g. This gives us the formula af (x) + bg(x) dx = a f (x) dx + b g(x) dx, which is analogous to the formula you already know for derivatives: d dx (af (x) + bg(x)) = a d dx f (x) + b d dx g(x).

7 Anti-Differentiation Techniques Example: Find 2x 2 sin(2x) dx. We know that the anti-derivative of x 2 is x3 3 and the anti-derivative of sin(2x) is 1 2 cos(2x), so 2x 2 sin(2x) dx = 2 3 x 3 ( 12 ) cos(2x) + C = 2 3 x cos(2x) + C. 2 There is one more technique commonly used in anti-differentiating that you will need to use. Example: Find cos 2 (x) dx.

8 Anti-Differentiation Techniques We don t know that anti-derivative for cos 2 (x), however we do know that cos 2 (x) = 1 2 (1 + cos(2x)), and so 1 cos 2 (x) dx = (1 + cos(2x)) dx 2 = cos(2x) dx 2 = 1 (x + 12 ) 2 sin(2x) + C = x sin(2x) + C. 4

9 Examples Now that we can evaluate anti-derivatives, we can answer two types of questions, the first are referred to as first order differential equations, where we have an equation involving the derivative of a function and we have to find the general solution. Example: Find the general solution to Solution: t dy = 1. dy = 1 t 1 y = = ln t + C. t

10 Examples The second are referred to as first order initial value problems, which consist of an equation involving the derivative of a function and the value of that function at some point. Example: Find y if t dy = 1, y(1) = 3. We know from before that y = ln t + C for some value of C. Knowing that y(1) = 3 gives us 3 = ln 1 + C = C, so y = ln t + 3.

11 First Order In most cases first order differential equations won t just involve t (or x), but also the original function itself. Example: Find the general solution to dy = yt. We take all expressions involving y to the left hand side, and all expressions involving t to the right and integrate. 1 dy = t, y 1 y dy = t ln y = t2 2 + C. Note that there s no point in putting a constant on both sides of the equals sign as we can just bring one to the other side and put them together.

12 First Order We then need to rewrite the equation to solve for y. y = e t2 2 +C = e C e t2 2 y = ±e C e t2 2 = Ce t2 2. Note that as the arbitrary constants aren t that important, we will often replace expressions like 2C or C with C, because fundamentally saying that the general solution is y = x + 2C and saying it is y = x + C is saying the same thing. Although it s a more complicated expression, replacing ±e C with C is doing the same thing. If you re uncomfortable with replacing C with itself you can use a new constant each time, but you often end up running out of constants quite quickly.

13 Second Order With this method we can solve a large variety of first order differential equations and initial value problems, although not all. For example, if you try to solve dy = y + t you ll find that you can t bring all the y and t terms to opposite sides. However you won t need to know how to solve such equations here. (For those interested, the type of equations that you now know how to solve are called separable, the rest are inseparable). We will now move on to second order differential equations; that is, equation that involve the second derivative of a function. There are two types of second order differential equations that you will see in this section: involving second and first derivatives (and maybe t), ( ) ex. d2 y = t dy 2. 2 involving second derivatives and the function itself, ex. d 2 y 2 = y.

14 Second Order : Type 1 ( ) = dy 2. 2 Notice that these are really just first order differential equations in We ll look at the first type now, for the equation d2 y disguise, if we let z = dy then dz = d2 y 2 and d2 y 2 = becomes dz = z2. We then solve this as normal: z 2 dz = z 2 dz = ( dy ) 2 z 1 = t + C 1 z = t + C z = 1 C t dy = 1 C t y = ln C t + D.

15 Second Order : Type 1 It is true in general that solutions to second order differential equations will have two arbitrary constants. Therefore initial value problems will give information about both y and dy at some specified times. Example: Solve the initial value problem d 2 ( ) y dy 2 2 = dy, = 1 and y = 2 when t = 0. Following the previous calculations would get us as far as dy = 1 C t, 1 = 1 C 0 1 = C dy = 1 1 t y = ln 1 t + D 2 = D y = ln 1 t + 2.

16 Second Order : Type 2 To solve the second type of second order differential equations, we will only deal with those that include initial value problems. We start by again trying to reduce the equation to a first order one, but in a slightly different way. Consider the equation in the example: d2 y 2 t = 0. Like before we let z = dy notice that dz = dz dy = y, with the initial values dy dy = dz dy = 2, y = 0 when dz, so that = d2 y. Now we 2 z. Therefore in this case we have dz dy z = y z dz = y dy z2 2 = y2 2 + C z 2 = y 2 + C z = ± y 2 + C dy = ± y 2 + C.

17 Second Order : Type 2 Now we have a first order differential equation to solve (unlike in the other example where we just had an anti-derivative to find). We can use the initial conditions now to find C, and to figure out the ± sign. Notice first that when t = 0, y 2 + C = C, which is non-negative. So if dy = 2, the ± sign must be a +, so we have 2 = C 4 = C. We therefore need to solve the initial value problem dy = 4 y 2, y(0) = 0. Soling this as normal, 1 4 y 2 dy = ( sin 1 y ) = t + D 2

18 Second Order : Type 2 y 2 = sin(t + D) y = 2 sin(t + D). Knowing that y = 0 when t = 0 gives us 0 = sin(d) 0, π = D y = 2 sin(t), 2 sin(t + D). To decide which is the real solution, we go back to the condition dy = 2 when t = 0 and see that as we get dy = 2 cos(t), 2 cos(t + π) = 2, 2 when t = 0, y = 2 sin t is the correct solution.

19 Examples of We now move on to some real-world problems which become differential equations when we try to solve them. The common theme here is that in these problems the rate of change of a quantity depends on the quantity itself (or something slightly more complex for second order problems). Example: The rate of change of the population of a town is proportional to the number of people living in that town. If 1000 people lived in this town on Jan 1st 2015, and 1200 lived here on Jan 1st 2016, how many will live in this city on Jan 1st 2017? We start by defining terms and solving the differential equation, we ll worry about the questions asked later. Let y be the population at time t where y is measured in Number of People and t is measured in years, where t = 0 is Jan 1st What we were told was that dy was proportional to y, so we start with dy = ky, y(0) = Then we solve the differential equation:

20 Examples of 1 y dy = k ln y = kt + C y = e kt+c y = ±e C e kt = Ce kt. Using the initial conditions, we get 1000 = C y = 1000e kt. We were asked for the value of y at time t = 2, but to do that we need to find k. That s where the knowledge that y(1) = 1200 comes in.

21 Examples of 1200 = 1000e k ln 1.2 = k k y = 1000e t y(2) = 1000e So there are approximately 1440 people living in this town on Jan 1st 2017.

22 Examples of Example: The engine of a car applies a constant force of 80 N. However there is also air resistance proportional to the velocity of the car. If the car is of mass m, what is the top speed of the car? If F is the sum of all forces acting on the car, what we were told was that F = 80 kv for some constant k. As F = ma and a = dv, we can rewrite this as a differential equation we can solve: m k m dv = 80 kv m 80 kv dv = ln 80 kv = t + C ln 80 kv = kt m + C 80 kv = e kt m +C 80 kv = ±e C e kt m v = 80 k + Ce kt m 80 k as t.

23 Examples of Example: An object lying on a table is connected to a point o on the table by a short elastic string. If the object is pulled a distance s from the table the string applies a force proportional to s in the direction of o. If the object is pulled 3 m away from o at time t = 0 and released, what is its position at time t? What we were told was that F = ks for some constant k, if we think of s as displacement rather than distance. As F = ma where m is the mass of the object and a = d2 s, we have the initial 2 value problem m d 2 s 2 = ks, By letting v = ds, we get d2 s 2 ds s = 3 and = dv = dv ds ds = 0 when t = 0. = dv ds m dv ds v = ks mv dv = ks ds v so that m v 2 2 = k s2 2 + C

24 Examples of v = ± C k m s2. The initial conditions give us 0 = ± C 9 k m which tell us that C = 9 k m but not what to do with the ± sign, so we ll leave it for now. We then have the differential equation ds = ± 9 k m ks2 m m 1 k ds = ± 9 s 2 m ( s ) k sin 1 = ±t + D 3 ( ) k s = 3 sin ± m t + D.

25 Examples of Using our initial conditions again we get that so that 3 = 3 sin(d), π 2 = D, ( ) k s = 3 sin ± m t + π. 2 You can then notice that whatever we replace the ± sign with, the function is the same. So by letting ± = + for convenience we get ( ) k s = 3 sin m t + π. 2