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Unformatted text preview: NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA 1506 Mathematics II Academic Year 2007-2008, Semester 2 7XWRULDO 1RWHV ZLWK 6ROXWLRQV Engineering Mathematics II: ............. JIN Chenyuan: Room S9A #02-03 Departments of Mathematics National University of Singapore 2 Science Drive 2, Singapore 117543 Republic of Singapore Office Number: 6516-8974 Email: [email protected] Homepage: http://jin.chenyuan.googlepages.com/ MA1506 Resources: http://jin.chenyuan.googlepages.com/ma1506 Acknowledgements This is the slides and notes for my tutorial groups of MA1505. I’d be grateful for any comments or corrections. I would like to thank . . . My thanks also goes out to the Department of Mathematics January 2008 Chapter 1 Tutorial 01 (Solution Notes) 1.1 Question 1 Solve the following differential equations: (a) x(x + 1)y = 1 (c) y = e(x−3y) (b) (sec(x))y = cos(5x) (d) (1 + y )y + (1 − 2x)y 2 = 0 3UHOLPLQDU\ • Separable Equations If an O.D.E. is of the form dy = M (x), dx then, we solve it with the following three steps: N (y ) 1. First, we can separate dy and dx as follows: N (y )dy = M (x)dx. 2. The integrate on both sides, we have N (y )dy = M (x)dx + c, where c is a constant. 3. (Sometimes this step is ignored.) Then try to solve out the explicit formula of y from the above equation y = ······ . 1 MA1506 Tutorial 01 (Solution Notes) 2 Remarks: 1. This is the method we have learnt in MA1505. 2. Although this solution only contains three steps (separation, integration, and explicitness), the difficulties may occur in each step. Actually, many solutions used in the lecture notes and tutorial problems are just trying to separate variables for special equations. So the solutions are not very hard. 3. If N (y ), the coefficient of y , is a nonzero constant, for example, N (y ) = 1, then life would be easier, we can omit the third step mentioned above. Check question (a) and (b). • Graphmatica This is one of the efficient software to sketch mathematical graphs. You can download a free trial version from http://www.graphmatica.com. (one-month free trial) But to sketch the graph of a given differential equation, we need some initial conditions. Otherwise, the solution has a constant c, which is a family of curves instead of a single one. For example in we have a initial condition y (1) = 0.5, then we can find the value of c, and sketch the curve with the command “x(x + 1)dy = 1 {1, 1/2}”. \  [  SSS  S S S S S S S S S S S S S S S    Remarks: 1. Sketch the curve of a given function BY HAND is required. That means in the exam, you cannot use the software or a calculator to help you. 2. The skill to get help from a software (MatLab, Maple, Mathematica, Graphmatica, etc.) is important in your Post-MA1506 career. You can learn the basic operations of MatLab during the lab section of MA1506. MA1506 Tutorial 01 (Solution Notes) 3 6ROXWLRQ (a) x(x + 1)y = 1 We can change the equation to y= 1 . x(x + 1) Therefore, this is a separable O.D.E., so we can solve the equation by integrating on both sides. Notice that 1 1 1 =− , x(x + 1) x x+1 which is called partial fraction, one of the basic tricks used in Calculus. Then we can easily get y= 1 dx − x 1 x dx + c = ln |x| − ln |x + 1| + c = ln + c. x+1 x+1 Remarks: 1. If you are not familiar with / have forgotten “partial fraction”, try to solve the following problems. Expanding Quotients into Partial Fractions 19. Expand the quotients in Exercises 1–8 by partial fractions. 1. 5x - 13 s x - 3 ds x - 2 d 2. 5x - 7 x 2 - 3x + 2 3. x+4 s x + 1 d2 4. 2x + 2 x 2 - 2x + 1 5. z+1 z 2s z - 1 d 6. z z 3 - z 2 - 6z 21. 7. t2 + 8 t 2 - 5t + 6 8. t4 + 9 t 4 + 9t 2 23. 27. x+4 11. dx 2 L x + 5x - 6 8 y dy 13. 2 4 L y - 2y - 3 15. dt 3 2 L t + t - 2t dx 2 L x + 2x 2x + 1 12. dx 2 L x - 7x + 12 1 y+4 dy 14. 2 1 L>2 y + y 16. x+3 dx 3 L 2x - 8x 28. 1 17. x 3 dx 0 L x + 2x + 1 2 dx 2 0 L s x + 1 ds x + 1 d y 2 + 2y + 1 L x 3 dx L 1 x - 2x + 1 2 dy 2s + 2 ds 2 3 L s s + 1 ds s - 1 d 24. 26. L L 1 L 3t 2 + t + 4 dt t3 + t L 8x 2 + 8x + 2 dx s 4x 2 + 1 d2 s 4 + 81 ds 2 2 L ss s + 9 d 2u3 + 5u2 + 8u + 4 du s u2 + 2u + 2 d2 u4 - 4u3 + 2u2 - 3u + 1 du s u2 + 1 d3 In Exercises 29–34, perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral. 29. 31. L 2x 3 - 2x 2 + 1 dx x2 - x 30. x4 dx 2 Lx -1 L 9x 3 - 3x + 1 dx x3 - x2 32. 16x 3 dx 2 L 4x - 4x + 1 y4 + y2 - 1 0 18. s y 2 + 1 d2 22. Improper Fractions Repeated Linear Factors In Exercises 17–20, express the integrands as a sum of partial fractions and evaluate the integrals. 23 1 In Exercises 9–16, express the integrands as a sum of partial fractions and evaluate the integrals. 10. x 2 dx s x - 1 ds x 2 + 2x + 1 d L In Exercises 21–28, express the integrands as a sum of partial fractions and evaluate the integrals. 25. dx 2 L1-x 20. Irreducible Quadratic Factors Nonrepeated Linear Factors 9. dx 2 2 L sx - 1d 33. L y +y 3 dy 34. 2y 4 2 Ly -y +y-1 3 Detailed solutions can be found in the solution book of Thomas’ Calculus (section 8.3): http://cid-... 2. The coefficient of y is 1, so the third step is omitted. dy MA1506 Tutorial 01 (Solution Notes) 4 (b) (sec(x))y = cos(5x) 1 , we can change the equation to Since sec x = cos x y = cos x cos 5x. Therefore, this is a separable O.D.E., so we can solve the equation by integrating on both sides. We have two method to deal with the integration on the right-hand-side: • By Trigonometric Identity Product-to-Sum Identity: cos A cos B = 1 cos(A + B ) + cos(A − B ) . 2 Therefore, cos x cos 5x = 1 cos 6x + cos 4x . 2 It follows that y= 1 2 cos 6x dx + cos 4x dx + c = 1 sin 6x sin 4x + +c 2 6 4 • By Integration by Parts Recall the formula: uv dx = uv − u v dx. Therefore, y= cos x cos 5x dx = sin x cos 5x − (−5) = sin x cos 5x + 5 sin x sin 5x dx (Int. by parts) sin x sin 5x dx = sin x cos 5x + 5 cos x sin 5x − 5 cos x cos 5x dx (Int. by parts) = sin x cos 5x + 5 cos x sin 5x − 5y = sin x cos 5x + 5 cos x sin 5x − 25y and so y + 25y = sin x cos 5x + 5 cos x sin 5x =⇒ y = 1 sin x cos 5x + 5 cos x sin 5x + c. 26 The answer is equivalent to the answer solving by identities. MA1506 Tutorial 01 (Solution Notes) 5 Remarks: 1. Check more identities from: http://en.wikipedia.org/wiki/List of trigonometric identities 2. You can use the online calculator to verify indefinite integral: http://integrals.wolfram.com/index.jsp (c) y = e(x−3y) We can change the equation to y= ex e3y =⇒ e3y dy = ex dx. Therefore, this is a separable O.D.E., so we can solve the equation by integrating on both sides: 1 3y e = ex + c. 3 This is enough, and you can write out the explicit solution easily by moving the coefficient 1 to the right-hand-side and taking logarithmic function ln on both sides. 3 (d) (1 + y )y + (1 − 2x)y 2 = 0 We can change the equation to (1 + y )y = (2x − 1)y 2 =⇒ 1+y y = (2x − 1) =⇒ y2 1 1 +2 yy dy = (2x − 1) dx. Therefore, this is a separable O.D.E., so we can solve the equation by integrating on both sides: ln |y | − 1 = x2 − x + c. y It is difficult to write out the explicit solution, so we stop here. &RPPHQWV 1. For constant coefficient (Qn (a) and (b)), we integrate the equation y = ··· . 2. For variable coefficient (Qn (c) and (d)), we integrate the equation · · · dy = · · · dx. 2 MA1506 Tutorial 01 (Solution Notes) 6 1.2 Question 2 Experiments show that the rate of change of the temperature of a small iron ball is proportional to the difference between its temperature T (t) and that of its environment, Tenv (which is constant). Write down a differential equation describing this situation. Show that T = Tenv is a solution. Does this make sense? The ball is heated to 300◦ F and then left to cool in a room at 75◦ F. Its temperature falls to 200◦ F in half an hour. Show that its temperature will be 81.6◦ F after 3 hours of cooling. 3UHOLPLQDU\ Mathematical Modeling From now on, we will see many word problems in the tutorial, how can we quickly solve these problems? Generally speaking, there are two steps: 1. Find a relation, and express the relation by a mathematical equation with variable asked in the problem (in this module, usually a differential equation); 2. Solve the equation, and find out the solution. For example, in the above question, the relation is from the first sentence: The rate of change of the temperature of a small iron ball is proportional to the difference between its temperature T (t) and that of its environment, Tenv (which is constant). We are asked what is the temperature at some time, and the temperature is denoted by a function T (t), which means the temperature at time t is T . Now, we know that – The rate of change of the temperature is expressed by the differentiation of temperature, that is dT . dt – y is proportional to x means there is a constant k, such that y = kx. Therefore, the first sentence means the differential equation dT = k(T − Tenv ). dt That is the model of the word problem. MA1506 Tutorial 01 (Solution Notes) 7 Remarks: 1. You need to have the modeling ability by practicing the tutorial problems and some extra problems. 2. One problem may have different models, and usually it is hard to say which is the best. For example, in the official solution, the model is built to be dT = −k(T − Tenv ). dt Naturally, one may ask why we add a negative sign here. One reason is that the change is reducing, so the left-hand-side term dT /dt is negative. However, since k here is an arbitrary constant (positive or negative), you will see that it is safe to ignore the negative sign and get the same answer. 6ROXWLRQ We answer the questions one by one: • Write down a differential equation describing this situation. We know (check Preliminary) the model of the problem is given by dT = k(T − Tenv ), dt and we don’t suffer to consider whether k is positive or negative here. • Show that T = Tenv is a solution. Does this make sense? To verify it, we just need to substitute it into the model, and it is easily to see that both the left-hand-side and right-hand-side are equal to 0, i.e., it is a solution. This does make sense because they have the same temperature, and objects do not spontaneously become hotter or colder. The solution is a trivial solution, to find out a nontrivial solution, we assume T = Tenv . • The ball is heated to 300◦ F and then left to cool in a room at 75◦ F. Its temperature falls to 200◦ F in half an hour. Show that its temperature will be 81.6◦ F after 3 hours of cooling. This part gives more conditions: T (0) = 300, Tenv = 75, T (0.5) = 200. We need to show T (3) = 81.6. Actually, we are asked to solve the initial problem and find the value of k: ⎧ ⎨ dT = k(T − 75) dt ⎩ T (0) = 300, T (0.5) = 200 MA1506 Tutorial 01 (Solution Notes) 8 We first consider the equation dT = k(T − 75). dt Obviously, it is a separable O.D.E., and can be written as the form: 1 dT = k dt. T − 75 Integrate on both sides, we can get ln |T − 75| = kt + c. By the initial condition T (0) = 300, T (0.5) = 200, we know that T > 75, and so we can remove the modulus ln(T − 75) = kt + c. Then we can write out the temperature as T = 75 + ekt+c . Substitute the two initial conditions T (0) = 300, T (0.5) = 200, we can get 300 = 75 + e0+c 200 = 75 + e0.5k+c From the first one, we know ec = 225, and from the second one, we know k = 2 ln =⇒ ek = 125 5 = 2 ln = ln 225 9 5 9 5 9 2 2 =⇒ ekt = (ek )t = 5 9 2t =⇒ T (t) = 75 + ekt+c = 75 + ekt ec = 75 + 225 =⇒ T (3) = 75 + 225 5 9 6 5 9 2t ≈ 81.6. This shows that its temperature will be 81.6◦ F after 3 hours of cooling. 2 &RPPHQWV Some basic operations of exponential function and logarithmic function are useful for initial problems: 1. ex+y = ex ey , exy = (ex )y = (ey )x ; 2. ln(xy ) = ln x + ln y , ln x = ln x − ln y . y MA1506 Tutorial 01 (Solution Notes) 1.3 9 Question 3 In very dry regions, the phenomenon called Virga is very important because it can endanger aeroplanes. See http://en.wikipedia.org/wiki/Virga. Virga is rain in air that is so dry that the raindrops evaporate before they can reach the ground. Suppose that the volume of a raindrop is proportional to the 3/2 power of its surface area. [Why is this reasonable? Note: raindrops are not spherical, but let’s assume that they always have the same shape, no matter what their size may be.] Suppose that the rate of reduction of the volume of a raindrop is proportional to its surface area. [Why is this reasonable?] Find a formula for the amount of time it takes for a virga raindrop to evaporate completely, expressed in terms of the constants you introduced and the initial surface area of a raindrop. Check that the units of your formula are correct. Suppose somebody suggests that the rate of reduction of the volume of a raindrop is proportional to the square of the surface area. Argue that this cannot be correct. 3UHOLPLQDU\ • Simplification Usually, the mathematical model is too complicated to be solved out. So, we will propose some assumptions to simplify the model. After we make an assumption, we need to check whether it is reasonable or not. For this question, we suppose all the raindrops have the same shape. • Dimensional Analysis This is used widely in Physics and related areas. What is dimensional analysis? Your lecturer assume you have learnt it in your Physics module, and you can check: http://en.wikipedia.org/wiki/Dimensional Analysis. – Basic Knowledge In Physics, we know that each physical quantity can be expressed by mass, length, time, electric charge, and temperature with some physical formula. It follows that the units of the quantity can be expressed by the combinations of kg, m, s, C, and ◦ F. – Example For example, we have formula F = ma, and so the corresponding units are N = kg·(m/s2 ). We describe the units N (Newton) by the units of mass, length and time. – Application Dimensional analysis can help you to check whether you have written down the correct formula. For example, if you write the volume of a cube to be V = a2 , then we can check the units of the left-hand-side is m3 , and the units of the right-hand-side is m2 . So the formula does not make sense. MA1506 Tutorial 01 (Solution Notes) 10 6ROXWLRQ • Suppose that the volume of a raindrop is proportional to the 3/2 power of its surface area. Why is this reasonable? We use the tool of dimensional analysis. We know that the units of volume is m3 , and the units of surface area is m2 . Therefore the relationship between the volume and area are coming from the identity m3 = (m2 )3/2 =⇒ V = aA3/2 , where a is a constant without units. This is reasonable! Remarks: Some in-depth thoughts. 1. How do we know to make such an assumption? We glance at the problem now. The major task is to find the time to evaporate the raindrop completely. So we need to know the change rate of the volume (the speed of evaporation). Actually, the speed of evaporation depends on the area of surface and the temperature of the liquid (learnt in secondary school?). The change of the temperature is due to the friction, and friction depends on the frictional area (the surface). Thus, the speed of evaporation depends on the surface area. Based on these analysis, we need to find out a relationship between the volume and surface area of the raindrop, and then we use dimensional analysis to find out the assumption. 2. Why a is a constant? Yes, the coefficient can change, so we assume that they always have the same shape, no matter what their size may be. Once the shape is fixed, then the coefficient a is fixed. For different shapes of objects, we have different coefficients. 43 1 r and A = 4πr 2 , then a = √ ; for cube, 3 6π V = l3 and A = 6l2 , then a = 6−3/2 For example, for sphere, V = • Suppose that the rate of reduction of the volume of a raindrop is proportional to its surface area. Why is this reasonable? This is based on a common sense of Physics: evaporation takes place on the surface of a raindrop, so large surface should give rise to a faster rate. It’s Physics! If you cannot understand the reasons, just skip these steps. MA1506 Tutorial 01 (Solution Notes) 11 • Find a formula for the amount of time it takes for a virga raindrop to evaporate completely, expressed in terms of the constants you introduced and the initial surface area of a raindrop. With the above assumptions, we can build two equations: ⎧ 3 ⎨V = aA 2 ⎩ dV = −bA dt The second equation, we add a negative sign before the constant b, the reason is the same as question 2. We substitute the first equation into the second one, and we have 3 d aA 2 dt = −bA. This is the model we need with two initial conditions: A(0) = A0 , A(teva ) = 0. Note that a bigger raindrop of course needs longer time to evaporate, so we need to know the initial size of the raindrop, either its volume or its surface area. Here, we suppose the initial surface area is A0 , if we are given the initial volume V0 , it is easy to get the A0 by the first equation of the model. To solve this model and find out the value of teva , we first observe that it is a separable O.D.E., since it can be written as 3a 1 dA A2 = −bA, 2 dt which is a separable O.D.E., since 1 A− 2 dA = −2b dt. 3a We integrate on both sides, and get √ √ c −2b −b 2 A= t + c =⇒ t+ . A= 3a 3a 2 Now we substitute the initial conditions √ √ A0 −b A(0) = A0 =⇒ c = 2 A0 =⇒ t+ . A= 3a 2 √ −b 3a A0 A(teva ) = 0 =⇒ 0 = teva + =⇒ teva = A0 . 3a 2 b 3a To evaporate a raindrop with initial surface area A0 , the time we need is b A0 . MA1506 Tutorial 01 (Solution Notes) 12 • Check that the units of your formula are correct. To check the units of teva = 3a b A0 , we know that: – The units of a is from the first equation of the model 3 V = aA 2 . For the equation, the left-hand-side has units m3 , and the right-hand-side has units (m2 )3 =m3 , thus, a has no units. – The units of b is from the second equation of the model dV = −bA dt For the equation, the left-hand-side has units m3 /s, and the right-hand-side has units m2 , thus, the units of b is m/s. – The units of A0 is m2 . 3a Therefore, the units of b that is meaningful! √ A0 is m2 = s, i.e., the units of the time teva is second (s), m/s • Suppose somebody suggests that the rate of reduction of the volume of a raindrop is proportional to the square of the surface area. Argue that this cannot be correct. Suppose the rate of reduction of the volume of a raindrop is proportional to the square of the surface area, then we need to modify the second equation of the model: ⎧ 3 ⎨V = aA 2 ⎩ dV = −bA2 dt We substitute the first equation into the second one, and we have 3 d aA 2 dt = −bA2 . Two initial conditions: A(0) = A0 , A(teva ) = 0. To solve this model and find out the value of teva , we use the same procedure. First observe that it is a separable O.D.E., since it can be written as 3a 1 dA A2 = −bA2 , 2 dt MA1506 Tutorial 01 (Solution Notes) 13 which is a separable O.D.E., since 3 A− 2 dA = −2b dt. 3a We integrate on both sides, and get 1 −2A− 2 = √ −2b t + c =⇒ A= 3a 1 . c −2 b 3a t Now we substitute the initial conditions 2 A(0) = A0 =⇒ c = − √ A0 =⇒ √ A= b 3a t 1 . + √1 A 0 A(teva ) = 0 =⇒ teva = ∞. That means to evaporate a raindrop with initial surface area A0 , the time we need is infinity! That is impossible for virga! If you visit http://en.wikipedia.org/wiki/Virga, you will see the first sentence says that: “In meteorology, virga is an observable streak or shaft of precipitation that falls from a cloud but evaporates before reaching the ground. · · · ” So the evaporation time must be finite for virga, i.e., the speed cannot be proportional to square of the surface area. 2 &RPPHQWV From this question, you may know the procedure of mathematical modeling. • First of all, we need some assumptions to simplify the problem. For example, in this question, we suppose the evaporating speed is proportional to the surface area. • Then we can solve the model we built with some mathematical knowledge. MA1506 Tutorial 01 (Solution Notes) 14 • Finally, we need to verify the result of the model is reasonable! Here we need some knowledge from other areas. If we find the result is impossible, that means the assumption is not correct, we need to go back to the first step to modify the assumption again. The last part of Question 3 is such an example. You are not major in Mathematics, and when you learn the Mathematical modules, usually you need to know three kinds of skills: 1. Translate the problem in your own major into a mathematical problem. That is Mathematical Modeling. • If you are a businessman, you need to build a model to determine the prices of your commodities to maximize your profit. • If you are a manager, you need to build a model to determine and schedule the priority of your works to maximize the efficiency of you and your employee. • If you are a speculator, you need to build a model to forecast the opportunity just like weather forecast. • If you are a engineer, you will use many components in your labs. Each component (accessory, reagent, electronic component, etc) will have its own properties expressed by numbers, these numbers are from the experiments and computations by modeling. To use these components, you need to build a model to combine these things together to get a best performance. Usually, these things are done by experiences, but will be much better with a good mathematical model. 2. The second skill is some mathematical knowledge. For example, the ability to solve a differential equation. 3. The last one is the academic ability. You cannot learn all branches of mathematics in college, and there will be more and more new useful branches in the future. You need to learn them by yourself if you feel it is useful to solve your model. For example, there are two hot branches of applied mathematics in recent years: game theory and fuzzy mathematics. You will learn mathematical modeling in chapter 3, but still at a basic level. MA1506 Tutorial 01 (Solution Notes) 1.4 15 Question 4 Moths navigate at night by keeping a fixed angle between their velocity vector and the direction of the Moon or some bright star (see http://en.wikipedia.org/wiki/Moth). A certain moth flies near to a candle and mistakes it for the Moon. What will happen to the moth? 3UHOLPLQDU\ • Polar Coordinates In MA1505, we have seen polar coordinates many times. We know that there are two variables used in polar coordinates: the radius (r ) and the angle (θ ). If the region is bounded by constant radius or constant angle, then we use polar coordinates to solve it.   5  T         T  5       Radius is a constant    Angle is a constant Here, we use polar coordinates to solve the problem, because there is a angle constant! Refer to http://en.wikipedia.org/wiki/Polar coordinates, if you are not familiar with polar coordinates. MA1506 Tutorial 01 (Solution Notes) 16 6ROXWLRQ The moth flies in such a way that the angle Ψ remains constant at all limes, so we use polar coordinates. Let us recall that in polar coordinates, each point is expressed by the radius r and the angle θ , to build up the polar coordinates, we need to set a point as the original point. The candle in the question is stationary, so we choose the point to be the original. WDQJHQW OLQH DW SRLQW $ % U U T T M U & UT M $ U  T T 2 T &DQGOH 0RWK V )OLJKW 0RWLRQ This is a picture of the flight path. When Δθ is very small, we may know that: 1. OA//OB =⇒ − → ∠ABC ≈ ϕ, where ϕ is the angle between the direction OA and the tangent line at point A. 2. |OA| ≈ |OC | =⇒ |BC | = |OB | − |OC | ≈ |OB | − |OA| = (r + Δr ) − r = Δr 3. |AC | ≈ r Δθ (the length of a circular arc) MA1506 Tutorial 01 (Solution Notes) So, we build a equation in the triangle 17 ABC : r Δθ . Δr Since the equation holds when the Δθ or Δr is very small, we let Δr → 0, and get a differential tan ϕ ≈ model: dθ . dr Since we are given a condition that the angle ϕ is a constant, then the left-hand-side of the tan ϕ = r equation is a constant. Hence the differential equation is an separable O.D.E.: dr dθ = . r tan ϕ So we integrate on both sides and get (note that tan Ψ is a constant): ln |r | = θ · cot ϕ + c =⇒ |r | = eθ·cot ϕ · ec =⇒ r = eθ·cot ϕ · ec (exponential function is always positive) We suppose initial condition is r(0) = R, then =⇒ R = e0 · ec =⇒ ec = R =⇒ r = R · eθ·cot ϕ Thus, we get the solution to the model: r = R · eθ·cot ϕ . Now we study what will happen to the moth after she first saw the candle at the point (r = R, θ = 0). Since the R is a given constant, and so the distant r depends on the exponential function. The exponential r = R · eθ·cot ϕ can be written as r = R · eθ·cot ϕ = R · (ecot ϕ )θ = R · E θ , where E = ecot ϕ . Now, there are three cases (depending on the value of E ): 1. If E = ecot ϕ > 1 (i.e., 0 < ϕ < π ), then E θ and r are increasing functions. 2 In this case, r is increasing means the moth will fly away from the candle. 2. If E = ecot ϕ = 1 (i.e., ϕ = π ), then E θ and r are constant functions (spiral outward). 2 In this case, r is constant means the moth will fly around the candle forever. 3. If E = ecot ϕ < 1 (i.e., π 2 < ϕ < π ), then E θ and r are decreasing functions (circle). In this case, r is decreasing means the moth will fly closer and closer to the candle (spiral). 2 MA1506 Tutorial 01 (Solution Notes) 18 1.5 Question 5 In Cosmology, the ratio of the sizes of the Universe at two different times is measured by a function of time called the scale function, denoted a(t). What are the units of a(t)? The (first) Friedmann Equation (http://en.wikipedia.org/wiki/Friedmann equations) relates this function to the energy density of the Universe and to its spatial curvature. In a particular cosmological model, the Friedmann equation takes the form L2 a2 = a2 − ˙ 2 + 1, a2 where L is a positive constant, the dot denotes time differentiation, and the initial condition is a(0) = 1. What are the units of L? Show, without solving this equation, that the universe described by this model is never smaller than a certain minimum size. Now solve the equation and describe the history of this universe. 3UHOLPLQDU\ • Notation of Differentiation There are three kinds of notations for differentiation that are widely used nowadays, and will appear in the lectures, tutorials, and tests of MA1505 and MA1506 frequently: – Leibniz’s Notation: dy . dx – Newton’s Notation: y (x). ˙ – Lagrange’s Notation: y (x). It is no use to distinguish these notations, just keeping in mind that all of the three notations are the same. By an incomplete figures, there are more than ten kind of notations expressing the differentiation. For more information, please refer to: http://en.wikipedia.org/wiki/Derivative. My favorite notation is the last one, since it is easy to write, easy to type, and difficult to be left out. One disadvantage of the notations y and y is that when we use dimensional analysis, it is ˙ easy to make mistake: the units of y is not the units of y , instead, the units should be the units of y . the units of x MA1506 Tutorial 01 (Solution Notes) 19 • Complete the Square You must be very familiar with this part. Given any quadratic polynomial ax2 + bx + c, we can write the polynomial as this form: ax2 + bx + c = a x + b 2a 2 + 4ac − b2 4a Check more details from: http://en.wikipedia.org/wiki/Complete the square. Please pay special attention to the “A variation on the technique” part! $ YDULDWLRQ RQ WKH WHFKQLTXH $V FRQYHQWLRQDOO\ WDXJKW FRPSOHWLQJ WKH VTXDUH FRQVLVWV RI DGGLQJ WKH WKLUG WHUP Y  WR WR JHW D VTXDUH 7KHUH DUH DOVR FDVHV LQ ZKLFK RQH FDQ DGG WKH PLGGOH WHUP HLWKHU XY RU í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utorial 01 (Solution Notes) 20 • Hyperbolic Functions I know most you have not learnt these kinds of function before college, but AT LEAST you should know the following things: 1. sinh x = ex − e−x ex + e−x sinh x , cosh x = and tanh x = with their sketches. 2 2 cosh x 2. cosh2 x − sinh2 x = 1. Hyperbolic functions will appear in the lecture notes, tutorials, and exams of MA1506 for four to five times. Remember it, useful! 6XSSOHPHQW RSWL RQ • “Mnemonic for Hyperbolic Formulae” by G. Osborn. Hyperbolic functions are now so constantly used, that a brief mnemonic for their somewhat confusing formulae may not be unwelcome. In any Trigonometrical formula for θ , 2θ , 3θ , or θ and φ, after changing sin to sinh, cos to cosh, etc., change the sign of any term that contains (or implies) a product of sinhs, e.g. tanh θ tanh φ implies a product of sinhs, ∴ tanh(θ + φ) = tanh θ + tanh φ . 1 + tanh θ tanh φ sinh 3θ = 3 sinh θ + 4 sinh 3θ ; cosh θ − cosh φ = +2 sinh θ+φ θ−φ sinh ; 2 2 and so on. This rule would fail for terms of the 4th degree, but it covers everything that is likely to be required, and is very convenient or teaching purposes. These rules are called as Osborn’s rule. • Useful Information About Hyperbolic Functions 1 For a full information of the hyperbolic functions, you can refer to the Word file: Actually, the permanent delegate of all the hyperbolic functions are: sinh, cosh, and tanh. Here, I want to list and summarize their basic properties. 1 For other hyperbolic functions, please check: “http://www.mediafire.com/?e0nrpxnmezj”. MA1506 Tutorial 01 (Solution Notes) 21 f sinh cosh tanh Definition of f e x − e −x 2 e x + e −x 2 e x − e −x e x + e −x parity Odd Even Odd Range of f −∞ < x < +∞ 1 ≤ x < +∞ −1 < x < 1 f (0) 0 1 0 f cosh(x) sinh(x) 1 cosh2 (x) coshx sinhx ln(cosh x) arcsinh(x) √ = ln(x + x2 + 1) arccosh(x) √ = ln(x ± x2 − 1) arctanh(x) √1 x 2 −1 1 1−x2 Domain of f −1 √1 x2 +1 −∞ < x < +∞ x≥1 −1 ≤ x ≤ 1 f −1 (0) 0 undefined 0 f dx f −1 (f −1 ) 1+x 1−x = ln • Relationship cosh2 (x) − sinh2 (x) = 1; ex = cosh(x) + sinh(x); e−x = cosh(x) − sinh(x); Trigonometric identities have already been stated above as Osborn’s rule. • Graphics MA1506 Tutorial 01 (Solution Notes) 22 You may notice that sinh and cosh are nearly the same when x is bigger than 2. Why? Think about this for a while, and see the following pictures, where the curves have been flattening. This is the flattening curve for sinh, where is cosh? The answer is above the curve of y= ex . 2 These graphics imply that ex = lim sinh(x), x → +∞ 2 x → +∞ lim sinh(x) = lim x → +∞ which can be easily proved by the definitions of sinh and cosh. To draw the graphics of a function, you can use MatLab, which you will learn in your lab section. For other hyperbolic functions, please check: http://en.wikipedia.org/wiki/Hyperbolic function. MA1506 Tutorial 01 (Solution Notes) 23 6ROXWLRQ The model of the problem is given: ˙ L2 a2 = a2 − 2 + 1, a2 with a initial condition a(0) = 1. Now we answer the questions one by one: • What are the units of a(t)? The scale function a(t) is the ratio of the sizes of the Universe at two different times, and so a(t) has no units. (If it is a ratio, then we will cancel all the units on denominator and numerator.) • What are the units of L? We find the units of L from the equation of the model. The right-hand-side of the equation is expressed by a, so no units on the right hand side. So is the left hand side, i.e., (La)2 ˙ has no units. It is the same to say La has no units. ˙ da 1 Note that a is ˙ , and the units of a is , and so the units of L is second (s). ˙ dt s • Show, without solving this equation, that the universe described by this model is never smaller than a certain minimum size. Suppose the universe can be smaller at some time t, i.e., a(t) < 1. If a < 1, then the right hand side of the equation, a2 − 2 a2 + 1 is negative. This is because 2 >2 a2 ◦ a < 1 =⇒ a2 < 1 =⇒ a2 + 1 < 2 ◦ a < 1 =⇒ a2 < 1 =⇒ Thus, a2 + 1 < 2 < i.e., RHS = a2 − 2 a2 2 2 + 1 = (a2 + 1) − 2 < 0. 2 a a However, the left hand side is a square (La)2 , so it is always nonnegative, which implies ˙ the right hand side is nonnegative. We get a contradiction. Therefore, the scale function a(t) cannot less than 1, that means the universe will not decrease. The Expanded Universe! (http://en.wikipedia.org/wiki/Expanded Universe) MA1506 Tutorial 01 (Solution Notes) 24 • Now solve the equation and describe the history of this universe. 2 ˙ The equation is L2 a2 = a2 − 2 + 1, with a initial condition a(0) = 1. a We note that the equation is second order! and if we take square root on both side, the right hand side will be a complicated function, and difficult to integrate! We need to change variable here. If you do not feel comfortable about the method, do the three equations in Question 6 first, then come back to this equation. Since a2 is a common part on the right hand side, so we do a substitution y = a2 , then da dy = 2a dt dt y = a2 =⇒ Then, we substitute da 1 dy = ·. dt 2a dt 1 dy da = · we can get: dt 2a dt L2 a2 = L2 ˙ Now, =⇒ da dt 2 = L2 1 dy · 2a dt 2 = L2 · 4a2 dy dt 2 = L2 · y2. ˙ 4a2 2 L2 2 + 1 =⇒ · y 2 = a2 − 2 + 1 ˙ 2 2 a 4a a 2 L 2 4 =⇒ · y 2 = y − + 1 =⇒ y 2 = 2 · (y 2 + y − 2) ˙ ˙ 4y y L 2 1 2 dy = dt y 2 + y − 2 =⇒ =⇒ y = ˙ L L y2 + y − 2 L2 a2 = a2 − ˙ This is a separable O.D.E., and we integrate on both sides, and get: 1 y2 + y − 2 dy = 2 dt =⇒ L 1 y2 + y − 2 Our next task is to evaluate the integral on the left hand side dy = 2 t L 1 y2 +y−2 (1.1) dy : For this integration, we use the following formula √ 1 x + c. dx = arccosh 2 a −a x2 The formula can be found in the formula list of Kreyszig’s book: Advanced Engineering Mathematics (9th Edition), I will attach the list in the last part of this question. Remark: This reason we know to choose this formula is because the denominator. The denominator of the integrand is a square root of a quadratic polynomial, and there are √ √ three formulas in the list may suit the case with denominator a2 − x2 , x2 + a2 , √ and x2 − a2 . We don’t know which formula will be used, but after we complete the square, there is only one formula we can apply. This trick is very useful for the later chapter “Laplace Transform”. MA1506 Tutorial 01 (Solution Notes) 25 To apply the formula, we need to change the quadratic polynomial in the square root to the form x2 − a2 , that is called “complete the square”. Just keep in mind that you want to apply the formula above, then every step and substitution is obvious. 1 y2 + y − 2 dy = = 1 (y + 0.5)2 − 1.52 1 (y + 0.5)2 − 1.52 1 √ = dv 2 − 1.52 v v +c = arccosh 1.5 dy (Complete the square) d(y + 0.5) (Change variable y → y + 0.5) (Substitution v = y + 0.5) Therefore, we solved the integral of the equation (1.1), and get arccosh v 2 + c = t. 1.5 L Note that, with the initial condition “y = 1 when t = 0”, we know: v = y + 0.5 = 1.5 =⇒ arccosh =⇒ 0 + c = We can write the solution as arccosh 1.5 = 0 (By the definition of cosh) 1.5 2 · 0 =⇒ c = 0. L 2 v = t. 1.5 L We take the operation “cosh” on both sides, we can get v = cosh 1.5 2t L . Since v = y + 0.5, then y + 0.5 = cosh 1.5 and so y= 3 cosh 2 2t L 2t L , 1 −. 2 With the graph of cosh x (check the picture on page 21), you can sketch y easily, and you will see that: a Universe that begins with nonzero size and then expands. The expansion is not slowing down, instead it is getting faster and faster. &RPPHQWV The following formula list may still useful in MA1506. 2 26 MA1506 Tutorial 01 (Solution Notes) MA1506 Tutorial 01 (Solution Notes) 1.6 27 Question 6 Solve the following equations: 1 − 2y − 4x (b) y = 1 + y + 2x (c) x + y + 1 + (−x + y − 3)y = 0 (a) y = x+y+1 x+y+3 2 3UHOLPLQDU\ • Substitution to Reduce O.D.E. to Be Separable Sometimes, the given ordinary differential equation is not separable. Then we can try to change the variable to reduce the equation to be a separable one. There is not unique rule to determine how we can change the variable, but just some tricks. Suppose the equation is of the form y = f (x, y ), then – When there is a common part appear in the function f (x, y ) on the right hand side, then we define the common part to be the new variable. For example, ∗ In Question 5, the common part is a2 , so we let y = a2 ; ∗ In Question 6(a), the common part is 2x + y , so we let v = 2x + y ; ∗ In Question 6(b), the common part is x + y , so we let Y = x + y – When we cannot find a common part, but similar part, usually we need to change two variables x and y . If we do linear substitution, we need to use the method called “Method of Undetermined Coefficients”, Question 6(c) is an example. Check details from: http://en.wikipedia.org/wiki/Undetermined coefficients method. – Most of the equations cannot be solved by substitution. Generally speaking, if the two-variable function f (x, y ), the right-hand-side of the equation y = f (x, y ), behaves as a single-variable function g( ), then we do a substitution v = . If you still remember the first problem of MA1505 tutorial 11, the function u(x, y ) = F (y − 3x) can be regarded as a single-variable function, and the solution is doing like that. Examples: * y = g(x + 4y ) =⇒ v = x + 4y * y = g(3x + 2y + 5) =⇒ v = 3x + 2y + 5 x x * y = g( ) =⇒ v = (We will also see this example in tutorial 2.) y y MA1506 Tutorial 01 (Solution Notes) 28 6ROXWLRQ These arc examples of O.D.E.s where a change of variable is needed. 1 − 2y − 4x 1 + y + 2x The equation can be written as (a) y = y= 1 − 2(2x + y ) . 1 + (2x + y ) The common part is 2x + y , so we do a substitution v = 2x + y , then v = 2 + y . From the substitution, we know that: ⎫ =⇒ y = v − 2 ⎪ ⎪ ⎬ v =2+y v = 2x + y y= =⇒ 1−2(2x+y ) 1+(2x+y ) y= 1−2v 1+v ⎪ ⎪ ⎭ =⇒ v −2 = 3 1 − 2v 1 − 2v =⇒ v = 2+ = . 1+v 1+v 1+v We get a new differential equation v= 3 , 1+v which is a separable O.D.E., since we can written the equation as (1 + v ) dv = 3 dx. Integrate on both sides, we have v+ v2 = 3x + c. 2 Finally, substitute v = 2x + y back, we can get an implicit solution: 1 (2x + y ) + (2x + y )2 = 3x + c. 2 (b) y = x+y+1 x+y+3 2 The common part is x + y , so we do a substitution v = x + y , then v = 1 + y . From the substitution, we know that: v =1+y v =x+y y= x+y +1 x+y +3 2 =⇒ v = 1 + ⎫ =⇒ y = v − 1 ⎬ 2 =⇒ y = v+1 v+3 ⎭ v+1 v+3 2 = ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ =⇒ v − 1 = v+1 v+3 v 2 + 4v + 5 2v 2 + 8v + 10 =2· 2 . v 2 + 6v + 9 v + 6v + 9 2 MA1506 Tutorial 01 (Solution Notes) 29 We get a new differential equation v =2· v 2 + 4v + 5 , v 2 + 6v + 9 which is a separable O.D.E., since we can written the equation as v 2 + 6v + 9 dv = 2 dx. v 2 + 4v + 5 Change the left hand side to be a proper fraction: 1+2· 2v + 4 dv = 2 dx. v 2 + 4v + 5 Integrate on both sides, we have v + ln v 2 + 4v + 5 = 2x + c. Finally, substitute v = 2x + y back, we can get an implicit solution: (x + y ) + ln (x + y )2 + 4(x + y ) + 5 = 2x + c. (c) x + y + 1 + (−x + y − 3)y = 0 Well, there is no common part for this equation, even we write it in the form: y= x+y+1 . x−y+3 Here we use another important trick to reduce the equation to a homogeneous equation. Remark: Homogeneity is an important notation in Mathematics, Physics and Statistics. You will see that the first chapter of MA1506 divide the O.D.E. into two parts: homogeneous and non-homogeneous, and solve them using different methods. The first-order homogeneous differential equation means the equation can be written as the form y y = f( ) x For example, if the equation is given by y= x+y , x−y then we can divide x on the denominator and numerator of the right-hand-side, and get y= 1 + y/x . 1 − y/x MA1506 Tutorial 01 (Solution Notes) 30 We may notice that in the original equation, each terms in the polynomials on the right-hand-side are all of the same order! That is reason why this is called homogeneous. Check your lecture notes or the page on wikipedia.org: http://en.wikipedia.org/wiki/Homogeneous (mathematics). We do a substitution here with undetermined coefficients α and β : x= X +α y =Y +β , where α and β are both constants. Note that: dx = d(X + α) = dX (by constant rule of Calculus) dy = d(Y + β ) = dY =⇒ dY dy x+y+1 = = dX dx x−y+3 (X + α) + (Y + β ) + 1 = (X + α) − (Y + β ) + 3 X + Y + (α + β + 1) . = X − Y + (α + β + 3) (substitution) Then, the best choice of α and β is satisfying (make the equation homogeneous) α+β+1=0 α−β+3=0 α = −2 =⇒ β=1 Thus, the substitution is: x=X −2 y =Y +1 or X =x+2 Y =y−1 . Using the substitution, we change the O.D.E. to a homogeneous equation: dY X +Y 1 + Y /X = = . dX X −Y 1 − Y /X The last equality is getting by divide X on denominator and numerator at the same time. Thus, the common part of the right-hand-side of the new O.D.E. is Y /X , and so we do the substitution Y . X We apply the product rule to Y = X · V , and can get V= Y = V + XV , MA1506 Tutorial 01 (Solution Notes) and so 31 ⎫ Y = V + XV ⎬ 1+V =⇒ V + XV = 1+V ⎭ 1−V Y= 1−V =⇒ (1 − V ) · (V + XV ) = 1 + V =⇒ V + XV − V 2 − XV V = 1 + V =⇒ (X − XV )V + (V − V 2 ) = 1 + V 1+V2 =⇒ V = X − XV 1+V2 1 dV = · =⇒ dX 1−V X 1 1−V dV = =⇒ dX 2 1+V X The last equation show that the new O.D.E. is a separable equation! Thus, we integrate on both sides: 1−V dV = 1+V2 1 dX. X The integral on the right-hand-side is easy, we can directly apply the formula, and get: 1 dX = ln |X | + c = ln |x + 2| + c. X For the left-hand-side, the integral can be solve in the following way: 1−V dV 1+V2 = 1 dV − 1+V2 V dV 1+V2 V dV (Check the formula list on page 25) 1+V2 1 arctan V − dV 2 (Due to Chain Rule: dV 2 = 2V dV ) 2(1 + V 2 ) 1 1 d(1 + V 2 ) (Due to Constant Rule) arctan V − · 2 1+V2 1 arctan V − · ln |1 + V 2 | 2 1 y−1 2 y−1 Y y−1 − · ln 1 + = V= arctan x+2 2 x+2 X x+2 = arctan V − = = = = The solution to the original equation is given by arctan y−1 1 y−1 − · ln 1 + x+2 2 x+2 2 = ln |x + 2| + c. MA1506 Tutorial 01 (Solution Notes) 32 1.7 Practice Questions The following are extracted from the Kreyszig’s book: Advanced Engineering Mathematics (9th Edition), They fit the tutorial very well. Some problems refer to the examples of section 1.3. MA1506 Tutorial 01 (Solution Notes) 33 Answers to Odd-Numbered Problems You can find the Even-Numbered solution from the book: http://cid-.... &RPPHQWV You have experienced suffering by MA1505, and you may realize that if you do not practise enough by yourself, you will not have time to think and write during the examination. Chapter 2 Tutorial 02 (Solution Notes) 2.1 Question 1 Solve the following differential equations: (a) xy + (1 + x)y = e−x , x>0 3 )y = x + 2, x x (c) y + y + = 0 y (b) y − (1 + y (1) = e − 1, (d) 2xyy + (x − 1)y 2 = x2 ex , x>0 x>0 3UHOLPLQDU\ This tutorial still consider first-order O.D.E., one is linear equations, the other is reducible non-linear equations. • First of all, let us recall there are three kinds of O.D.E.s solved in tutorial one: 1. Separable Equation N (y ) dy = M (x). dx 2. Reducible Equation I: y = f (ax + by + c) 3. Reducible Equation II: y =f y x Refer to tutorial one for more details. 35 MA1506 Tutorial 02 (Solution Notes) 36 • Integrating Fator 1 If a linear first-order O.D.E. is of the form dy + P (x)y = Q(x). dx Then we are trying to define a integrating factor R(x) = e x a P (s)ds . Since R = RP by the chain rule, so (Ry ) = RP y + Ry . Now, multiply both sides of the original O.D.E. by R(x), we get Ry + RP y = RQ, or (Ry ) = RQ. Integrate on both sides then we have R(x)y = R(x)Q(x)dx + c, which means R(x)Q(x)dx + c y= . R(x) Note that this solution type is suit for many kinds of O.D.E.s (though may not be the best method), then this is one of the most important methods. • Linear-Reducible Type One - Bernoulli Equation 2 Bernoulli Equation of the form dy + p(x)y = q (x)y n , dx is a special case that is not linear but can reduced to a linear form. Here n = 1, otherwise, this equation can be solved by integrating factor. The solution steps are: first define z = y 1−n , then dz (1 − n)dy = , dx y n dx and the Bernoulli equation is dz + (1 − n)p(x)z = (1 − n)q (x). dx Now, the equation is in linear form, and use the integrating factor R(x) = e(1−n) 1 x a p(s)ds . This method is founded by Daniel Bernoulli, who was a Swiss mathematician who spent much of his life in Basel where he died. A member of a talented family of mathematicians, physicists and philosophers, he is particularly remembered for his applications of mathematics to mechanics, especially fluid mechanics, and for his pioneering work in probability and statistics. 2 If you are interested in Physics, Mechanics, Hydraulics, and Fluid Dynamics, you can learn something about Bernoulli’s Principle by referring to the page: http://en.wikipedia.org/wiki/Bernoulli’s principle. MA1506 Tutorial 02 (Solution Notes) 37 6ROXWLRQ (a) xy + (1 + x)y = e−x , x>0 This is a linear equation, since it can be written in the form y + 1+ 1 x y= 1 −x e, x and so we can use integrating factor to solve it. The integrating factor is given by exp 1+ 1 x = ex+ln |x| = ex · eln |x| = |x| · ex = xex . We can remove the modulus in the last equality since we assume x > 0 in the equation. Then, we multiply the integrating factor xex on both sides of the equation, and get y xex + 1 + 1 x y xex = By chain rule, the left-hand-side is equal to 1 −x x e xe . x d (yxex ), and the right-hand-side is 1, i.e., dx d (yxex ) = 1. dx It is easy to get yxex = x + c, and so y = e−x + cx−1 e−x . 3 )y = x + 2, y (1) = e − 1, x > 0 x This is a linear equation, and so we can use integrating factor to solve it. (b) y − (1 + The integrating factor is given by (don’t forget to add the negative sign in the integrand) exp − 1+ 3 x = e−x−3 ln |x| = e−x · e−3 ln |x| = |x|−3 · e−x = x−3 e−x . We can remove the modulus in the last equality since we assume x > 0 in the equation. Then, we multiply the integrating factor x−3 e−x on both sides of the equation, and get y x−3 e−x − 1 + 3 x y x−3 e−x = (x + 2)x−3 e−x . By chain rule, the left-hand-side is equal to d (yx−3 e−x ), and so dx d (yx−3 e−x ) = (x + 2)x−3 e−x . dx MA1506 Tutorial 02 (Solution Notes) 38 Integrate on both sides, we can get ye−x = x3 e−x dx + 2 x2 e−x dx + c. x3 Note that with integration by parts, we have e−x dx = − x3 2 e−x d 1 1 = − e−x · 2 − x2 x 1 de−x x2 =− e−x − x2 e−x dx. x2 Thus, 2 e−x e−x dx = − 2 − x3 x e−x dx =⇒ x2 e−x dx + 2 x2 e−x e−x dx = − 2 . x3 x This happens to be the integrals on the right-hand-side, so substitute it to the solution of the equation, we get moving the coefficient ye−x e−x = − 2 + c, x3 x e−x of y to the right-hand-side, x3 y = −x + cx3 ex . With the initial condition y (1) = e − 1, we have y (1) = (−x + cx3 ex ) x=1 = −1 + c · 13 · e1 = −1 + ce = e − 1 =⇒ c = 1. Therefore, the solution to the initial problem is y = −x + x3 ex . x =0 y This is NOT a linear equation, but Bernoulli Equation (n = −1, p(x) = 1, q (x) = −x) (c) y + y + dy + p(x)y = q (x)y n , dx so we can reduce it to linear form and use integrating factor to solve it. Step 1 Do substitution z = y 1−n . In this case, n = −1, and so the substitution is z = y 2 Step 2 Solve the linear equation z + (1 − n)p(x)z = (1 − n)q (x) by its integrating factor. In this case, n = −1, p(x) = 1, q (x) = −x, and so the linear equation is z + 2z = −2x. The integrating factor is R=e p(x) dx =e 2 dx = e2x , MA1506 Tutorial 02 (Solution Notes) 39 so we consider the differential equation integrate on both sides, we have d 2x d (Rz ) = Rq (x), i.e., (e z ) = e2x · (−2x). dx dx 1 ze2x = (−x + )e2x + c. 2 Thus, 1 − x + ce−2x . 2 Remember the substitution we did is z = y 2 , so the solution to the differential equaz= tion is 1 − x + ce−2x . 2 y2 = (d) 2xyy + (x − 1)y 2 = x2 ex , x>0 This is NOT a linear equation, and the coefficient of y contains y . So, first, we need to do some substitution to remove the coefficient. Note that 2yy = (y 2 ) (chain rule), then the differential equation can be written as x(y 2 ) + (x − 1)y 2 = x2 ex . So, we do the substitution Y = y 2 , and have a new linear differential equation xY + (x − 1)Y = x2 ex . Divide x on both sides (to remove the coefficients of Y ): Y + 1− 1 x Y = xex . We use integrating factor to solve this linear equation: R = exp p(x) dx 1 x = exp(x − ln |x|) = exp 1− dx = exp(x − ln x) (x > 0 is assumed in the equation) x e = x Now, consider the equation (Rz ) = Rq (x), which is d dx ex Y x = ex · xex . x Integrate on both sides, we have ex 1 Y = e2x + c, x 2 so the solution is Y = y2 = 1x xe + cxe−x . 2 2 MA1506 Tutorial 02 (Solution Notes) 40 2.2 Question 2 If a cable is held up at two ends at the same height, then it will sag in the middle, making a U-shaped curve called a catenary. This is the shape seen in electricity cables suspended between poles, in countries less advanced than Singapore, such as Japan and the US. It can be shown using simple physics that if the shape is given by a function y (x), then this function satisfies x dy μ = dx T 2 dy dt 0 + 1 dt, where x = 0 at the lowest point of the catenary and y (0) = 0, where μ is the weight per unit length of the cable, and where T is the horizontal component of its tension; this horizontal component is a constant along the cable. Find a formula for the shape of the cable. 6ROXWLRQ The differential equation is given in the question: x dy μ = dx T 2 dy dt 0 + 1 dt, with initial condition y (0) = 0, and substituting x = 0 into the equality above will get y (0) = 0. dy , then The common part is the derivative of y , and so we define a substitution v = dx v= μ T x 0 dy dt 2 + 1 dt. By Fundamental Theorem of Calculus, dv μd = dx T dx x 0 dy dt 2 + 1 dt = μ T dy dx 2 +1= μ T v 2 + 1, i.e., dv μ = dx T v 2 + 1. This is a separable equation, since we can write it as 1 μ √ dv = dx. 2+1 T v We integrate on both sides, and can get 1 √ dv = 2+1 v μ dx. T μ x + c. To evaluate the left-handT side integral, we need to use the most important hyperbolic identity The integral on the right-hand-side is easy, which is equal to cosh2 x − sinh2 x = 1. MA1506 Tutorial 02 (Solution Notes) 41 Remarks: 1. If there is a term 2. If there is a term √ √ 1 − v 2 , then we use the identity sin2 x + cos2 x = 1. 1 + v 2 , then we use the identity cosh2 x − sinh2 x = 1. We use the substitution, v = sinh w, and so 1 1 √ d sinh w dv = 2+1 v sinh2 w + 1 1 √ = cosh w dw cosh2 w 1 = cosh w dw | cosh w| 1 cosh w dw = cosh w = (sinh w) = cosh w cosh w = ew + e−w >0 2 1 dw = w Thus, the solution to the separable equation is μ w = x + c, T and so μ x+c . v = sinh w = sinh T Now consider the initial condition v (0) = y (0) = 0, we have μ ec − e−c · 0 + c = sinh c = = 0 =⇒ ec = e−c =⇒ e2c = 1 =⇒ c = 0. T 2 μ dy Therefore, v = sinh x . We know the substitution is v = , and so T dx μ dy = sinh x. dx T We integrate on both sides, and get T μ y = cosh x + c. μ T With the initial condition, we know T y (0) = 0 =⇒ y (0) = cosh ·0 +c μ T T cosh 0 + c = μ T e0 + e−0 = +c cosh 0 = =1 μ 2 =0 (Initial condition.) T =⇒ c=− μ μ T T x− . 2 Hence, the solution to the initial problem is y = cosh μ T μ v (0) = sinh MA1506 Tutorial 02 (Solution Notes) 42 2.3 Question 3 Psychologists talk about something called a Performance Curve. Suppose an MA1506 student is solving mathematics problems. She starts with differential equations. Let P (t) be a nonnegative function that measures her performance, that is, her success rate at solving D.E.s. Her performance increases rapidly at first, but then the rate of increase slows down as she becomes more expert. Let M, a positive constant, be the best possible performance; then one can suppose that P satisfies dP = C [M − P ], dt where C is a constant. What are the units of this constant? What does this constant measure? Solve this equation assuming that she is completely incompetent at t = 0 [that is, P (0) = 0]. Now the student turns to another kind of problem, say in linear algebra. Again her performance is low at first but gets better in accordance with this equation. Now as the years go by, her overall ability to solve mathematics problems gradually gets better, so C , instead of being a constant, is really a slowly increasing function of time. Suppose that C (t) = K tanh(t/T ), t ≥ 0, where K and T are positive constants. Is this reasonable? Why? What are the meanings of the constants K and T ? Replacing C with C (t), solve for P , again assuming that P (0) = 0. Sketch P (t), the performance curve. (Choose values of the constants for yourself, and use Graphmatica if necessary.) 6ROXWLRQ The differential equation is given in the question dP = C [M − P ], dt where C is a constant. We answer the questions one by one. • What are the units of this constant? – Since P is the rate of her success, so P has no units. Thus, the units of the left-handdP 1 side is . dt time – The units of both sides should be the same, thus, the units of the right-hand-side is 1 . time – Moreover, M is the performance, so M has no units, which implies M − P has no units. Thus, the units of the right-hand-side has the same units as the constant C . Hence, the units of the constant C should be 1 . time MA1506 Tutorial 02 (Solution Notes) 43 • What does this constant measure? The constant C measures how fast P is increasing. • Solve this equation assuming that she is completely incompetent at t = 0 (that is, P (0) = 0). dP = C [M − P ] is separable, since we can write the equation as The differential equation dt 1 dP = C dt. M −P Integrate on both sides, we can get 1 1 dP = − d(−P ) M −P M −P 1 =− d(M − P ) M −P (Add two “-” to the integral) (Constant Rule) = − ln M − P = − ln(M − P ) (M is the best performance, so M > P.) C dt = Ct Thus, the solution to the differential equation is given by − ln(M − P ) = Ct + D, where D is a constant. Since the initial condition says P = 0 when t = 0, we have − ln M = D . Thus, − ln(M − P ) = Ct − ln M. This implies (by taking exponential function on both sides) P = M − M e−Ct . • Suppose that C (t) = K tanh(t/T ), t ≥ 0, where K and T are positive constants. Is this reasonable? Why? First, let us recall the graph of y = tanh x is \    [                    MA1506 Tutorial 02 (Solution Notes) 44 The graph of C (t) looks similar to the curve in the first quadrant, and C (t) measures the rate of increase of the performance P . We would expect the following: 1. C (t) is bounded (clearly everyone has a limitation) 2. At the start, when you are fresh into a subject, you will probably pick up the elementary parts quite fast. Thus, C (t) increases quite fast at first 3. As time goes on, you will be done with the elementary parts and get into the deeper parts which cannot be picked up as fast. Thus, C (t) increases less fast after a while. We can see that the graph of C (t) = K tanh(t/T ) fits the three conditions. y .5 1 K .5 x 0 O0 0.5 1 T 1.5 2 2T 2.5 33 T 3.5 Thus, the assumption is reasonable. • What are the meanings of the constants K and T ? t ≤ K . Thus, K is the least upper T bound of the function C (t). In other words, K is the maximum possible speed of – Since tanh x ≤ 1, we have C (t) = K tanh learning. – T is a cut off point. ∗ Before T , since you will be mainly picking up the elementary parts, you learn quite fast, i.e., C (t) increases fast before T . ∗ After T , since you will then be doing the deeper stuff, your learning will slow down, i.e., C (t) will not be increasing as fast as before. Thus, T means the amount of time required for her to realize her maximum potential. MA1506 Tutorial 02 (Solution Notes) 45 • Replacing C with C (t), solve for P , again assuming that P (0) = 0. In this case, the equation is t T dP = K tanh dt [M − P ]. This is a linear equation, since it can be written as dP + K tanh dt t T t T P = K tanh M. Then the integrating factor is t dt T sinh(t/T ) = exp K dt cosh(t/T ) sinh(t/T ) = exp K T d(t/T ) cosh(t/T ) 1 d cosh(t/T ) = exp K T cosh(t/T ) R = exp K tanh (Multiply T and 1/T to the integral) = exp K T ln cosh(t/T ) = exp K T ln cosh(t/T ) = exp ln coshKT (t/T ) = coshKT (t/T ) By the formula of integrating factor (check the preliminary of question one), we know the solution to the differential equation is given by P = = = = = = = 1 R R · KM tanh 1 KT dt (KM tanh coshKT t T t T tanh t T · KM tanh coshKT t T cosh KM t coshKT T KM t coshKT T KM T t coshKT T KM T t coshKT T KM T t coshKT T t T t T dt dt coshKT −1 t T sinh t T dt coshKT −1 t T sinh t T d coshKT −1 · t T t T 1 coshKT KT = M + cKM T sechKT t T d cosh t T +c t T t T is the term on the RHS.) (Substitute R) MA1506 Tutorial 02 (Solution Notes) 46 Note that the initial condition is given by P (0) = 0, and so P (0) = M + cKM T sechKT 0 = M + cKM T = 0 =⇒ c = − 1 . KT Hence, the solution to the initial problem is P (t) = M − M sechKT t T • Sketch P (t), the performance curve. (Choose values of the constants for yourself, and use Graphmatica if necessary.) If we choose M = 2, K = 3, T = 6, then use the command is “y = 2−2/(cosh(x/6))18 {0, 6}”. Here is the example of the sketch: \      [     &RPPHQWV              MA1506 Tutorial 02 (Solution Notes) 2.4 47 Question 4 A certain MA1506 student starts a rumour to the effect that one of the lecturers has been seen dating Tang Wei. The number of students who have heard the rumour, R(t), is given by dR = KR[1500 − R], dt where K is a positive constant, and 1500 is the number of students taking MA1506. What is the meaning of K ? Is this equation reasonable? [Hint: surely the rumour will spread slowly both when hardly anyone has heard it yet, but also when nearly everyone has already heard it!] By regarding this equation as a Bernoulli equation, find R(t). Note that, in reality, R(t) is actually an integer [whole number]. Comment on this fact in relation to your solution. 6ROXWLRQ The differential equation is given by dR = KR[1500 − R], dt where K is a positive constant, and 1500 is the number of students taking MA1506. We answer the questions one by one: • What is the meaning of K ? K measures the rate of increase of R. It will probably depend on how juicy the rumour is (the juicier the rumour, the larger the value of K becomes); how much the students like to gossip, how gullible they are, etc. • Is this equation reasonable? [Hint: ] Notice that the range of R is from 1 to 1500. Surely the rumour will spread slowly both when hardly anyone has heard it yet (near R = 1), but also when nearly everyone has already heard it (R = 1500)! • By regarding this equation as a Bernoulli equation, find R(t). We write the equation in the following form dR − 1500KR = −KR2 . dt dy + p(x)y = q (x)y n as discussed in the notes, where n = 2. dx By the solution steps, we define a substitution Z = R1−n = R−1 , and so This is a Bernoulli equation dR dZ = − R −2 , dt dt MA1506 Tutorial 02 (Solution Notes) 48 i.e., dZ dR = −R 2 . dt dt Thus, we substitute into the differential equation, we have −R 2 dZ − 1500KR = −KR2 , dt divide −R2 on both sides, and substitute R−1 by Z , then we have a linear equation, (Dividing by -R2 ) (Substitution) dZ + 1500KR−1 = K dt dZ + 1500KZ = K dt The integrating factor of this linear equation is exp 1500K dt = e1500Kt . Now, the solution of the linear equation is given by 1 Z= e1500Kt = e−1500Kt e1500Kt · K dt 1 1500Kt e +c 1500 1 + ce−1500Kt 1500 = Since we do the substitution Z = R−1 , we have 1 1 = + ce−1500Kt . R 1500 Notice that the initial condition is R(0) = 1 (the rumour was started by one student), then R(0) = 1 1499 1 + ce0 = + c = 1 =⇒ c = . 1500 1500 1500 Therefore, the solution to the initial problem is R= 1 1499 −1500Kt + e 1500 1500 −1 • Note that, in reality, R(t) is actually an integer. Comment on this fact in relation to your solution. Of course as t tends to infinity, R tends to 1500. The function R(t) is not really continuous or differentiable, since it only takes integer values. However, this integer-valued function can be well approximated by a smooth function defined to interpolate between integral values. 2 MA1506 Tutorial 02 (Solution Notes) 2.5 2.5.1 49 Practice Questions Exercise Set 1 The following are extracted from the Kreyszig’s book: Advanced Engineering Mathematics (9th Edition), They fit the tutorial very well. Some problems refer to the examples of section 1.4. MA1506 Tutorial 02 (Solution Notes) 50 Remark: Here an O.D.E. of the form M (x, y ) + N (x, y )y = 0 is called an exact O.D.E., if the vector field M (x, y ), N (x, y ) T is a conservative field. For these equations, we can find their potential function to solve it. Answers to Odd-Numbered Problems You can find the Even-Numbered solution from the book: http://cid-.... 2.5.2 Exercise Set 2 The following are extracted from the Kreyszig’s book: Advanced Engineering Mathematics (9th Edition), They fit the tutorial very well. Some problems refer to the examples of section 1.5. MA1506 Tutorial 02 (Solution Notes) 51 52 MA1506 Tutorial 02 (Solution Notes) MA1506 Tutorial 02 (Solution Notes) Answers to Odd-Numbered Problems You can find the Even-Numbered solution from the book: http://cid-.... 53 Chapter 3 Tutorial 03 (Solution Notes) 3.1 Question 1 Solve the following differential equations: (a) y + 6y + 9y = 0, y (0) = 1, (b) y − 2y + (1 + 4π 2 )y = 0, y (0) = −1 y (0) = −2, y (0) = 2(3π − 1) 3UHOLPLQDU\ • Homogeneous Equations with Constant Coefficients The general form of homogeneous linear second-order O.D.E. with constant coefficients is y + ay + by = 0. Remark: 1. Homogeneous means the right-hand-side is 0. But the equation y + 2y + x = 0 is not homogeneous. (Be careful with these equations.) 2. Linear means we do not have a term like y 2 or y 3 or y 4 , etc. 3. Second-order means the equation contains y but no higher-order derivatives like y . 4. If the equation is of the form ay + by + cy = 0, then we can divide a on both sides, and get the general form. 5. You do not need to remember the long name, just remember the general form is enough. 55 MA1506 Tutorial 03 (Solution Notes) 56 These O.D.E.s can be solved as follows: Consider the corresponding characteristic equation λ2 + aλ + b = 0, which has two roots λ1 = 1 (−a + 2 a2 − 4b) and 1 λ2 = (−a − 2 a2 − 4b). Then there are three cases: – If a2 − 4b > 0, then λ1 and λ2 are both real numbers, and the general solution is of the form: y = Aeλ1 x + Beλ2 x . – If a2 − 4b = 0, then λ1 = λ2 = − a , and the general solution is of the form: 2 y = (A + Bx)e− ax 2 . – If a2 − 4b < 0, then λ1 and λ2 are both complex numbers, and the general solution is of the form: y = e− where ω = b− ax 2 (A cos ωx + B sin ωx), a2 . 4 6ROXWLRQ (a) y + 6y + 9y = 0, y (0) = 1, y (0) = −1 The corresponding characteristic equation is λ2 + 6λ + 9 = 0, and the root of this equation is a double root: λ = 3. Thus, the solution is given by the second case: y = (A + Bx)e− ax 2 = (A + Bx)e−3x . MA1506 Tutorial 03 (Solution Notes) 57 To find the values of the coefficients, we consider the initial conditions: (A + Bx)e−3x • y (0) = 1 =⇒ • y (0) = −1 =⇒ x=0 = 1 =⇒ A = 1 B e−3x − 3(A + Bx)e−3x x=0 = B − 3A = −1 =⇒ B = 2 Therefore, the solution to the initial problem is given by: y = (1 + 2x)e−3x . (b) y − 2y + (1 + 4π 2 )y = 0, y (0) = −2, y (0) = 2(3π − 1) The corresponding characteristic equation is λ2 − 2λ + (1 + 4π 2 ) = 0, and the roots of this equation are complex: λ = 1 ± 2πi. Thus, the solution is given by the third case, and ω = y = e− ax 2 A cos ωx + B sin ωx b− a2 = 2π : 4 = ex A cos 2πx + B sin 2πx . To find the values of the coefficients, we consider the initial conditions: • y (0) = −2 =⇒ ex (A cos 2πx + B sin 2πx) • y (0) = 2(3π − 1) =⇒ x=0 = −2 =⇒ A = −2 y + ex (−2πA sin 2πx + 2πB cos 2πx) =⇒ y (0) + 2πB = 2(3π − 1) x=0 = 2(3π − 1) =⇒ −2 + 2πB = 2(3π − 1) =⇒ B = 3 Therefore, the solution to the initial problem is given by: y = ex − 2 cos 2πx + 3 sin 2πx . 2 &RPPHQWV Homogeneous equation is a special case of differential equations, but very important for solving non-homogeneous equations. MA1506 Tutorial 03 (Solution Notes) 58 3.2 Question 2 Find particular solutions of the following: (a) y + 2y + 10y = 25x2 + 3 (b) y − 6y + 8y = x2 e3x (c) y − y = 2x sin x (d) y + 4y = sin2 x 3UHOLPLQDU\ • Purpose: For nonhomogeneous equation, the major part is to find a particular solution. The reason can be seen from the following example, we use the equation y + 2y + 10y = 25x2 + 3 as an example: 1. Find the general solution to the corresponding homogeneous equation y + 2y + 10y = 0 with the method used in question 1. We suppose the solution is y = h(x). 2. Find a particular solution to the non-homogeneous equation y + 2y + 10y = 25x2 +3, saying y = g(x). 3. The general solution to the non-homogeneous equation y + 2y + 10y = 25x2 + 3 is given by y = h(x) + g(x). From the above steps, we know the first step can be easily solved like question one, and the last step is trivial. So the difficult part is the second step, find a particular solution to the non-homogeneous equation. • Model: The general form of non-homogeneous linear second-order O.D.E. with constant coefficients is y + ay + by = r (x). Obviously, the particular solution to the differential equation depends on the function r (x) on the right hand side. We try to find out the particular solution in the following preliminary. MA1506 Tutorial 03 (Solution Notes) 59 • Puzzle: It seems that if we choose the polynomial u(x) with the same degree as p(x), then the candidate of y will satisfy the equation. But sometimes, this fails. In the lecture notes, Example 19 gives a counterexample; in the tutorial, question 3.2 (d) also fails with the same degree. So, a natural question arises: how to choose the polynomial u(x) here. One treatment is like the lectue notes, just suppose u(x) here, then substitute the candidate of y into the differential equation, then change the equation to another equation, whose right-hand-side is a polynomial, then using the trick above to solve the problem. An alternative way is using some rule to choose some proper candidate. In the official solution to tutorial question 3.2(d), we deal in this way, but do not tell you why and how to do the choice like that. You can find the method with some examples in the tutorial comments later. • Method: MA1506 Tutorial 03 (Solution Notes) 60 The particular solution can be find by cases as follows: – If r (x) = Pm (x)eμx , where Pm (x) is a polynomial, and then try1 the function given by the following rule. Suppose r (x) = Pm (x)eμx , where Pm (x) is a polynomial with degree m, then the differential equation has a particular solution of the following form: y = xk Qm (x)eμx , where, ∗ Qm (x) is a polynomial with degree m: Qm (x) = Am xm + Am−1 xm−1 + · · · + A1 x + A0 , where all the coefficients are undetermined. ∗ k is the multiplicity of root μ in the characteristic equation λ2 + aλ + b = 0. (If μ is not a root of the equation, then k = 0.) Remarks: 1. Question (a) and (b) are using this rule. 2. If r (x) is a polynomial, it is the same if we multiply an e0x (for example, r (x) = x2 − 2x +3 = (x2 − 2x +3)e0x ), so the polynomial case can be regard as a special case of this part (μ = 0). This means, we just check whether 0 is a root of the corresponding characteristic equation. – r (x) is a trigonometric function sin(mx) or cos(mx), then replace r (x) with eimx , y with z , and consider the equation z + az + bz = eimx . Solve this equation using the trick above, and ∗ If r (x) = cos(mx), then y = Re(z ). ∗ If r (x) = sin(mx), then y = Im(z ). Remarks: Question (c) and (d) are in this case. 1 In the lecture notes, we usually try a polynomial with the same degree as r (x), and there is a danger that this polynomial may fail. But if you use the rule mentioned, the polynomial is actually the correct polynomial we need to solve the differential equation. MA1506 Tutorial 03 (Solution Notes) 61 Using the above candidate of y , we can calculate y and y , and substitute all of them into the equation, and get the actual values of the undetermined coefficients. • Example Here are an example of the rule mentioned above to help you understand the rule. We use question (a) as an example: First note that the right hand side is a polynomial, and we can rewrite the polynomial as r (x) = 25x2 + 3 = (25x2 + 3) · e0 . Compare with the rule r (x) = Pm (x)eμx , we know that – Pm (x) = 25x2 + 3 =⇒ Qm (x) = Ax2 + Bx + C (Second-Order Polynomial with Undetermined Coefficients) – μ=0 By the rule, we should try the function of the form xk Qm (x)eμx . Now we need to determine the value of k: 1. If the differential equation is y + 2y + 10y = 25x2 + 3, then the corresponding characteristic equation is λ2 + 2λ + 10 = 0. μ = 0 is not a root of the equation, so k = 0, and then we should try the function xk Qm (x)eμx = x0 (Ax2 + Bx + C )e0 = Ax2 + Bx + C . 2. If the differential equation is y + 2y = 25x2 + 3, then the corresponding characteristic equation is λ2 + 2λ = 0 with roots λ = 0 and λ = −2. μ = 0 is a root of the equation (multiplicity = 1), so k = 1, and then we should try the function xk Qm (x)eμx = x1 (Ax2 + Bx + C )e0 = Ax3 + Bx2 + Cx. 3. If the differential equation is y = 25x2 + 3, then the corresponding characteristic equation is λ2 = 0. μ = 0 is a double root of the equation (multiplicity = 2), so k = 2, and then we should try the function xk Qm (x)eμx = x2 (Ax2 + Bx + C )e0 = Ax4 + Bx3 + Cx2 . MA1506 Tutorial 03 (Solution Notes) 62 6ROXWLRQ Find particular solutions of the following: (a) y + 2y + 10y = 25x2 + 3 To find the particular solution, we try the polynomial y = Ax2 + Bx + C . Notes: We follow the rule mentioned in the preliminary: “For r (x) = Pm eμx , try y = xk Qm (x)eμx .” The right-hand-side is a polynomial with degree 2, and here μ = 0 is not a root of λ2 + 2λ + 10 = 0, i.e., k = 0, so we try the function y = x0 · (Ax2 + Bx + C ) · e0 = Ax2 + Bx + C. Substitute y = Ax2 + Bx + C into the differential equation y + 2y + 10y = 25x2 + 3, we get: (Note that y = Ax2 + Bx + C =⇒ y = 2Ax + B =⇒ y = 2A) y + 2y + 10y = 2A +2· 2Ax + B + 10 · Ax2 + Bx + C = 10Ax2 + (4A + 10B )x + (2A + 2B + 10C ) ⎧ ⎪ 10A = 25, ⎪ ⎨ =⇒ 4A + 10B = 0, ⎪ ⎪ ⎩ 2A + 2B + 10C = 3 =⇒ 5 A= , 2 B = −1, C = 0. 5 Thus, y = x2 − x is a particular solution to the equation. 2 (b) y − 6y + 8y = x2 e3x To find the particular solution, we try the function y = (Ax2 + Bx + C )e3x . Notes: We follow the rule mentioned in the preliminary: “For r (x) = Pm eμx , try y = xk Qm (x)eμx .” The polynomial contained in right-hand-side r (x) has degree 2, and here μ = 3 is not a root of λ2 − 6λ + 8 = 0, i.e., k = 0, so we try the function y = x0 · (Ax2 + Bx + C ) · e3x = (Ax2 + Bx + C )e3x . MA1506 Tutorial 03 (Solution Notes) 63 Note that by product rule, y = (Ax2 + Bx + C )e3x =⇒ y = (2Ax + B )e3x + 3(Ax2 + Bx + C )e3x =⇒ y = 2Ae3x + 3(2Ax + B )e3x + 3(Ax + B )e3x + 9(Ax2 + Bx + C )e3x Substitute y , y and y into the differential equation y − 6y + 8y = x2 e3x , we get: y − 6y + 8y = 2Ae3x + 3(2Ax + B )e3x + 3(Ax + B )e3x + 9(Ax2 + Bx + C )e3x −6 · = =⇒ =⇒ (2Ax + B )e3x + 3(Ax2 + Bx + C )e3x +8· (Ax2 + Bx + C )e3x (9A − 18A + 8A)x2 + (6A + 6A + 9B − 12A − 18B + 8B )x +(2A + 3B + 3B + 9C − 6B − 18C + 8C ) e3x ⎧ ⎪ 9A − 18A + 8A = 1, ⎪ ⎨ 6A + 6A + 9B − 12A − 18B + 8B = 0, ⎪ ⎪ ⎩ 2A + 3B + 3B + 9C − 6B − 18C + 8C = 0 A = −1, B = 0, C = −2. Thus, y = (−x2 − 2)e3x is a particular solution to the equation. (c) y − y = 2x sin x The right hand side contains a sine function, so we consider a complex differential equation instead: z − z = 2xeix To find the particular solution of this complex equation, we try the function z = (Ax + B )eix . Notes: We follow the rule mentioned in the preliminary: “For r (x) = Pm eμx , try z = xk Qm (x)eμx .” The polynomial contained in right-hand-side r (x) has degree 1, and here μ = i is not a root of λ2 − 1 = 0, i.e., k = 0, so we try the function z = x0 · (Ax + B ) · eix = (Ax + B )eix . Note that by product rule, z = (Ax + B )eix =⇒ z = Aeix + i(Ax + B )eix =⇒ z = 2iAeix − (Ax + B )eix MA1506 Tutorial 03 (Solution Notes) 64 Substitute z , z and z into the differential equation z − z = 2xeix , we get: z −z − = 2iAeix − (Ax + B )eix = (−2A)x + (2iA − 2B ) eix −2A = 2, =⇒ =⇒ (Ax + B )eix 2iA − 2B = 0 A = −1, B = −i. Thus, z = (Ax+B )eix = (−x−i)eix = (−x−i)(cos x+i sin x) = (sin x−x cos x)+i(− cos x−x sin x) is a particular solution to the complex equation. The original real differential equation contains sine function, so we take the imaginary part of z , i.e., y = Im(z ) = Im (sin x − x cos x) + i(− cos x − x sin x) = − cos x − x sin x, i.e., y = − cos x − x sin x is a particular solution to y − y = 2x sin x. (d) y + 4y = sin2 x Note that only first-order sine and cosine functions can change to exponential functions by Euler’s formula eix = cos x + i sin x. sin x −→ eix , cos x −→ eix . So we do the modification: r (x) = sin2 x = 11 − cos 2x. 22 Now, the right hand side contains a cosine function, so we consider a complex differential equation instead: z + 4z = 1 1 2ix −e. 22 To find the particular solution of this complex equation, we try the function z = A+Bxe2ix . Notes: We follow the rule mentioned in the preliminary: “For r (x) = Pm eμx , try z = xk Qm (x)eμx .” r (x) contains two part 1 1 and − e2ix , so the particular solution will have two parts: 2 2 MA1506 Tutorial 03 (Solution Notes) • 65 1 : 2 This part is a polynomial with degree 0, and here μ = 0 is not a root of λ2 + 4 = 0, i.e., k = 0, so we try the function z1 = x0 · (A) · e0 = A. 1 • − e2ix : 2 The polynomial contained in this part has degree 0, and here μ = 2i is a root of λ2 + 4 = 0 with multiplicity 1, i.e., k = 1, so we try the function z2 = x1 · (B ) · e2ix = Bxe2ix . Thus, we try the function: z = z1 + z2 = A + Bxe2ix . Note that by product rule, z = A + Bxe2ix =⇒ z = Be2ix + 2iBxe2ix =⇒ z = 4iBe2ix − 4Bxe2ix Substitute z , z and z into the differential equation z + 4z = 1 2 − 1 e2ix , we get: 2 z + 4z 4iBe2ix − 4Bxe2ix = = 4A =⇒ =⇒ +4· A + Bxe2ix − 4iB e2ix 4A = 1 , 2 4iB = − 1 2 1 1 A = , B = i. 8 8 Thus, 1 + 8 is a particular solution z = A + Bxe2ix = 1 2ix 1 1 ixe = (1 + ix(cos 2x + i sin 2x)) = (1 + x(i cos 2x − sin 2x)) 8 8 8 to the complex equation. The original real differential equation contains cosine function, so we take the real part of z , i.e., y = Re(z ) = Re i.e., y = 1 11 (1 + x(i cos 2x − sin 2x)) = − x sin 2x, 8 88 11 − x sin 2x is a particular solution to y − y = 2x sin x. 88 &RPPHQWV Please refer to the webcast: http://cid- 2 MA1506 Tutorial 03 (Solution Notes) 66 3.3 Question 3 Use the method of variation of parameters to find particular solutions of (a) y + 4y = sin2 x (b) y + y = sec x 3UHOLPLQDU\ • Variation of Parameters For the O.D.E. of the form y + a(x)y + b(x)y = r (x), suppose we have y1 and y2 as the basic solutions of the homogeneous problem y + a(x)y + b(x)y = 0, then the general solution of the homogeneous problem is yh = c1 y1 (x) + c2 y2 (x). Compute Wronskian2 W = y1 y2 − y1 y2 , and coefficients u(x) = − y2 r dx, W v (x) = y1 r dx. W Then the particular solution is yp = u(x)y1 (x) + v (x)y2 (x), and the general solution is y = yh + yp = c1 y1 (x) + c2 y2 (x) + u(x)y1 (x) + v (x)y2 (x). 2 A function named after the Polish mathematician J´zef Hoene - Wro´ ski. It is especially important in the o n study of differential equations. MA1506 Tutorial 03 (Solution Notes) 67 6ROXWLRQ (a) y + 4y = sin2 x 1. First find two solutions of the homogeneous problem. The homogeneous equation is y + 4y = 0. The corresponding characteristic equation is λ2 + 4 = 0, which has two complex root, so the general solution to the homogeneous one is given by yh = e− ax 2 c1 cos ωx + c2 sin ωx , where a = 0 and ω = 2, and so yh = c1 cos 2x + c2 sin 2x. Here we have two solutions y1 = cos 2x and y2 = sin 2x. 2. Variation of Parameters The Wronskian is given by W = y1 y2 − y1 y2 = cos 2x sin 2x − sin 2x cos 2x = cos 2x 2 cos 2x − sin 2x − 2 sin 2x = 2 cos2 2x + 2 sin2 2x = 2. Thus, the coefficients u(x) = − =− y2 r dx W sin 2x sin2 x dx 2 = = v (x) = = = = (b) y + y = sec x &RPPHQWV y1 r dx W (r is the right-hand-side of the O.D.E.) ...
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This note was uploaded on 12/09/2011 for the course ELECTRICAL 101 taught by Professor Huanhoang during the Fall '11 term at National University of Singapore.

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