Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble to derive the differentil eqution modeling phsicl or geometric sitution. 4. Be ble to solve seprble differentil eqution, including finding lost solutions. 5. Be ble to solve n initil vlue problem (IVP) b solving the differentil eqution nd using the initil condition to find the constnt of integrtion. 1.2 Differentil equtions nd solutions A differentil eqution (DE) is n eqution with derivtives! Emple 1.1. (DE s modeling phsicl processes, i.e., rte equtions) 1. Newton s lw of cooling: dt = k(t A), where T is the temperture of bod in n dt environment with mbient temperture A. 2. Grvit ner the erth s surfce: m d2 = mg, where is the height of mss m dt2 bove the surfce of the erth. 3. Hooke s lw: m d2 dt 2 = k, where is the displcement from equilibrium of spring with spring constnt k. Other emples: Below we will give some emples of differentil equtions modeling some geometric situtions. A solution to differentil eqution is n function tht stisfies the DE. Let s focus on wht this mens b contrsting it with solving n lgebric eqution. The unknown in n lgebric eqution, such s 2 + 2 + 1 = 0 is the number. The eqution is solved b finding numericl vlue for tht stisfies the eqution. You cn check b substitution tht = 1 is solution to the eqution shown. The unknown in the differentil eqution d 2 d 2 + 2 d d + = 0 1
1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 2 is the function ().The eqution is solved b finding function () tht stisfies the eqution One solution to the eqution shown is () = e. You cn check this b substituting () = e into the eqution. Agin, note tht the solution is function. More often we will s tht the solution is fmil of functions, e.g. = Ce t. The prmeter C is like the constnt of integrtion in 18.01. Ever vlue of C gives different function which solves the DE. 1.3 The most importnt differentil eqution in 18.03 Here, in the ver first clss, we stte nd give solutions to our most importnt differentil equtions. In this cse we will check the solutions b substitution. As we proceed in the course we will lern methods tht help us discover solutions to equtions. The most importnt DE we will stud is d dt =, (1) where is constnt (in units of 1/time). In words the eqution ss tht the rte of chnge of is proportionl to. Becuse of its importnce we will write down some other ws ou might see it: = ; d dt = (t); = 0;. = 0. In the lst eqution we used the phsicist dot nottion to indicte the derivtive is with respect to time. You should recognize tht ll of these re the sme eqution. The solution to this eqution is (t) = Ce t, where C is n constnt. 1.3.1 Checking the solution b substitution The bove solution is esil checked b substitution. Becuse this eqution is so importnt we show the detils. Substituting (t) = Ce t into Eqution 1 we hve: Left side of (1): Right side of (1): = Ce t = Ce t Since fter substitution the left side equls the right, we hve shown tht (t) = Ce t is indeed solution of Eqution 1.
1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 3 1.3.2 The phsicl model of the most importnt DE As phsicl model this eqution ss tht the quntit chnges t rte proportionl to. Becuse of the form the solution tkes we s tht Eqution 1 models eponentil growth or dec. In this course we will lern mn techniques for solving differentil equtions. We will test lmost ll of them on Eqution 1. You should of course understnd how to use these techniques to solve (1). However: whenever ou see this eqution ou should remind ourself tht it models eponentil growth or dec nd ou should know the solution without computtion. 1.4 The second most importnt differentil eqution Our second most importnt DE is m + k = 0, (2) where m nd k re constnts. You cn esil check tht, with ω = k/m, the function (t) = C 1 cos(ωt) + C 2 sin(ωt) is solution. Eqution 2 models simple hrmonic oscilltor. More prosicll, it models mss m oscillting t the end of spring with spring constnt k. 1.5 Solving differentil equtions b the method of optimism In our first nd second most importnt equtions bove we simpl told ou the solution. Once ou hve possible solution it is es to check it b substitution into the differentil eqution. We will cll this method, where ou guess solution nd check it b plugging our guess into the eqution, the method of optimism. In ll seriousness, this will be n importnt method for us. Of course, its utilit depends on lerning how to mke good guesses! 1.6 Generl form of differentil eqution We cn lws rerrnge differentil eqution so tht the right hnd side is 0. For emple, = cn be written s = 0. With this in mind the most generl form for differentil eqution is F (t,,,..., (n) ) = 0, where F is function. For emple, ( ) 2 + e sin(t) (4) = 0. The order of differentil eqution is the order of the highest derivtive tht occurs. So, the emple just bove shows DE of order 4.
1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 4 1.7 Constructing differentil eqution to model phsicl sitution We use rte equtions, i.e. differentil equtions, to model sstems tht undergo chnge. The following rgument using t should be somewht fmilir from clculus. Emple 1.2. Suppose popultion P (t) hs constnt birth nd deth rtes: β = 2%/er, δ = 1%/er Build differentil eqution tht models this sitution. nswer: In the intervl [t, t + t], the chnge in P is given b P = number of births - number of deths. Over smll time intervl t the popultion is roughl constnt so: Births in the time intervl P (t) β t Deths in the time intervl P (t) δ t Combining these we hve P (β δ)p (t). t Finll, letting t go to 0 we hve derived the differentil eqution dp dt = (β δ)p = kp. Notice tht if β > δ then the popultion is incresing. Of course, this DE is our most importnt DE (1): the eqution of eponentil growth or dec. We know the solution is P = P 0 e kt. Note: If β nd δ re more complicted nd depend on t, s β = P + 2t nd δ = P/t. The derivtion of the DE is the sme, but becuse β nd δ re no longer constnts this is not sitution of eponentil growth nd the solution will be more complicted (nd probbl hrder to find). Emple 1.3. Bcteri growth. Suppose popultion of bcteri is modeled b the eponentil growth eqution P = kp. Suppose tht the popultion doubles ever 3 hours. Find the growth constnt k. nswer: The eqution P = kp hs solution P (t) = Ce kt. From the initil condition we hve tht P (0) = C. Since the popultion doubles ever 3 hours we hve P (3) = Ce 3k = 2C. Solving for k we get k = 1 3 ln 2 (in units of 1/hours.) 1.8 Initil vlue problems An initil vlue problem (IVP) is just differentil eqution where one vlue of the solution function is specified. We illustrte with some simple emples. Emple 1.4. Initil vlue problem. Solve the IVP ẏ = 3, (0) = 7.
1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 5 nswer: We recognize this s n eponentil growth eqution, so (t) = Ce 3t. Using the initil condition we hve (0) = 7 = C. Therefore (t) = 7e 3t. Emple 1.5. Initil vlue problem. Solve the IVP = 2, (2) = 7. nswer: Note, the use of indictes tht the independent vrible in this problem is. This is rell n 18.01 problem: integrting we get = 3 /3+C. Using the initil condition we find C = 7 8/3. 1.9 Seprble Equtions Now it s time to lern our first technique for solving differentil equtions. A first order DE is clled seprble if the vribles cn be seprted from ech other. We illustrte with series of emples. Emple 1.6. Eponentil growth. Use seprtion of vribles to solve the eponentil growth eqution = k. nswer: We rewrite the eqution s d = 4. Net we seprte the vribles b getting dt ll the s on one side nd the t s on the other. d = 4 dt. Now we integrte both sides: d = 4 dt ln = 4t + C. Now we solve for b eponentiting both sides: = e C e 4t or = ±e C e 4t. Since ±e C is just constnt we renme it simpl K. We now hve the solution we knew we d get: = Ke 4t. Emple 1.7. Here is stndrd emple where the solution goes to infinit in finite time (i.e. the solutions blow up ). One of the fun fetures of differentil equtions is how ver simple equtions cn hve ver surprising behvior. Solve the initil vlue problem d dt = 2 ; (0) = 1. nswer: We cn seprte the vribles b moving ll the s to one side nd the t s to the other d 2 = dt Integrting both sides we get: 1 = t + C
1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 6 Think: The constnt of integrtion is importnt, but we onl need it on one side. Solving for we get the solution: = 1 t + C. Finll, we use initil condition (0) = 1 to find tht C = 1. (t) = 1 1 t. So the solution is: We grph this below. Note tht the grph of the solution hs verticl smptote t t = 1. = 1 1 t t 2 1 1 2 3 4 Grph of solution 1.9.1 Technicl definition of solution Looking t the previous emple we see the domin of consists of two intervls: (, 1) nd (1, ). For technicl resons we will require tht the domin consist of ectl one intervl. So the bove grph rell shows two solutions: Solution 1: (t) = 1/(1 t), where is in the intervl (, 1) Solution 2: (t) = 1/(1 t), where is in the intervl (1, ) In the emple problem, since our IVP hd (0) = 1 the solution must hve t = 0 in its domin. Therefore, solution 1 is the solution to the emple s IVP. 1.9.2 Lost solutions We hve to cover one more detil of seprble equtions. Sometimes solutions get lost nd hve to be recovered. This is smll detil, but ou wnt to p ttention since it s worth 1 es point on ems nd psets. Emple 1.8. In the emple = 2 we found the solution = 1. But it is es to + C check b substitution tht (t) = 0 is lso solution. Since this solution cn not be written s = 1/( + C) we cll it lost solution. The simple eplntion is tht it got lost when we divided b 2. After ll if = 0 it ws not legitimte to divide b 2. Generl ide of lost solutions for seprble DE s
1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 7 Suppose we hve the differentil eqution = f()g() If g( 0 ) = 0 then ou cn check b substitution the () = 0 is solution to the DE. It m get lost in when we seprte vribles becuse dividing b b g() would then men dividing b 0. Emple 1.9. Find ll the (possible) lost solutions of = ( 2)( 3). nswer: In this cse g() = ( 2)( 3). The lost solutions re found b finding ll the roots of g(). Tht is, the lost solutions re () = 2 nd () = 3. 1.9.3 Implicit solutions Sometimes solving for s function of is too hrd, so we don t! Emple 1.10. Implicit solutions. Solve = 3 +3+1 6 ++1. nswer: This is seprble nd fter seprtting vribles nd integrting we hve 7 7 + 2 2 + = 4 4 + 32 2 + + C. This is too hrd to solve for s function of so we leve our nswer in this implicit form. 1.9.4 More emples Emple 1.11. Solve d d =. nswer: Seprting vribles: d = d. Therefore d = d, which implies ln = 2 2 + C. Finll fter eponentition nd replcing ec b K we hve = Ke 2 /2. Think: There is lost solution tht ws found b some slopp lgebr. Cn ou spot the solution nd the slopp lgebr? Emple 1.12. Solve d d = 3 2. nswer: Seprting vribles nd integrting gives: 1 = 4 4 There is lso lost solution: () = 0. Emple 1.13. Solve + p() = 0. 4 = 4 + 4C. + C. Solving for we hve nswer: We first rewrite this so tht it s clerl seprble: d = p() d. After the usul seprtion nd integrtion we hve log( ) = p() d + C Therefore, () = e C e p() d nd () = 0 is lost solution.
1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 8 1.10 Geometric Applictions of DEs Since the slope of curve is given b its derivtives we cn often use differentil equtions to describe curves. Emple 1.14. An hev object is drgged through the snd b rope. Suppose the object strts t (0, ) with the puller t the origin, so the rope hs length. The puller moves long the -is so tht the rope is lws tut nd tngent to the curve followed b the object. This curve is clled trctri. Find n eqution for it. nswer: The digrm below shows tht d d = 2 2 (, ) 2 2 The trctri Thus 2 2 d = d. Integrting (detils below) we get ( + ) ln 2 2 2 2 = + C. ( + ) The initil position (, ) = (0, ) implies C = 0. Therefore = ln 2 2 2 2. To finish the problem we show tht the integrl is wht we climed it ws: Let I = 2 2 d. Now use the trig. substitution: = sin u: cos u cos 2 u I = cos u du = sin u sin u du 1 sin 2 u = du = csc u sin u du sin u = ln(csc u + cot u) cos u Bck substituting we get I = ( 2 2 + ) + ln 2 2, which is wht we climed bove.
1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 9 Emple 1.15. Suppose = () is curve in the first qudrnt nd tht the prt of the curve s tngent line tht lies in the first qudrnt is bisected b the point of tngenc. Find nd solve the DE for this curve. nswer: The figure shows the piece of the tngent bisected b the point (, ) on the curve. Thus, the slope of tngent = d d =. This differentil eqution is seprble nd is esil solved: = C/. 2 (, )