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MATH 4B–Differential Equations, Fall 2016Final Exam Study GuideGENERAL INFORMATION AND FINAL EXAM RULES The exam will have a duration of 3 hours. No extra time will be given. Failing tosubmit your solutions within 3 hours will result in your exam not being graded. The Final Exam is comprehensive. The sections are 1.1–1.3, 2.1–2.6, 3.1–3.7, 6.1–6.4,7.1–7.9. 35% of the questions will be from Chapters 1 and 2, 35% from Chapters 3 and 6, and30% from Chapter 7. You can bring ONE index card of dimensions up to 500 600 . This index card should behandwritten and can be filled on both sides. However, note cards of higher dimensionsthan the ones mentioned above or typewritten WILL NOT be allowed. Calculators WILL NOT be needed, nor allowed for this exam. Last but not least, CHEATING WILL NOT BE TOLERATED.1
SKILL’S LIST Verify that a given function is a solution for an Initial Value Problem (IVP). Sketch the direction (slope) fields of an ODE. Find equilibrium solutions for Autonomous ODEs, and determine whether equilibrium solutions are semistable or not. Solve a separable ODE and corresponding IVP. Find the general solution to a first order linear ODE. Determine if an ODE is exact. Solve an exact ODE. Read information from a word problem, and establish the corresponding ODE modelingthe situation in the following cases:1. Free falling object.2. Population growth and decay.3. Tank model.4. Newton’s law of cooling.5. Springs. Determine whether two functions y1 (t), y2 (t) form a fundamental set of solutions fora second order linear ODE. Find the general solution for a second order homogeneous ODE with constant coefficients. Given a solution y1 (t) for a second order linear ODE, use reduction of order to find asecond (independent) solution of the form y2 (t) u(t)y1 (t). Find a particular solution for a second order ODE with constant coefficients using themethod of undetermined coefficients, and/or variation of parameters. Find the general solution to a second order ODE with constant coefficients. Solve Initial Value Problems associated to a second order ODE with constant coefficients.2
Compute the Laplace transform of a given function. Compute inverse Laplace transforms to rational functions and piecewise continuousfunctions. Use Laplace transforms to solve second order IVP. Compute the determinant of a matrix (2 2 or 3 3 would suffice). Find the eigenvalues and eigenvectors of a matrix. Find the inverse of a matrix. Find the canonical Jordan form of a 2 2 matrix. Find the general solution of a homogeneous system of linear ODEs with constantcoefficients. Find the fundamental matrix of a system of linear ODEs with constant coefficients. Find the solution to IVPs of the form:x0 Ax, x(t0 ) x0 . Find the general solution to a 2 2 system of linear ODEs with constant matrix A.Must be familiar with:1. Jordan form method,2. Undetermined coefficients,3. Variation of parameters, and4. Laplace transform.3
SUMMARY An Ordinary Differential Equation (ODE) is an equation whose variable is a differentiable function and that involves the function and its derivatives:F (x, y, y 0 , y 00 , . . . , y (n) , . . . ) 0. The order of an ODE is the order of the highest derivative that appears in the equation.For instance, y 0 y x is first order while x3 y 00 10x7 y 0 y 4 is second order. A first order ODE of the formy 0 f (y)is called autonomous. Notice that autonomous equations are always separable, however the integration involved may be very complicated (if not impossible). One canin any case study solutions by looking at the direction (slope) field for the differentialequation. A constant solution to an Autonomous ODE is called an equilibrium solution. Theseare the zeroes of the function f (y). An equilibrium solution is called semistable ifas t approaches infinity solutions on one side of the equilibrium solution approach theequilibrium solution while solutions on the other side become further and further awayfrom the equilibrium solution; it is called asymptotically stable if for large valuesof t solutions always approach the equilibrium solution, in population dynamics thiswould say that no matter the initial size of a population it will (after a long time)stabilize to the equilibrium size; it is called unstable if for large values of t everysolution stays away from the equilibrium. There are special kinds of ODEs that we know how to solve, these are:1. Separable: First order ODEs of the formy 0 F (t)G(y).These can be solved by dividing both sides by G(y) and then integrating withrespect to t:ZZy 0 (t)dt F (t) dt,G(y(t))the left hand side is actually the integral with respect to y since if y y(t) thendy y 0 (t)dt, and so the equation becomesZZ1dy F (t) dt.G(y)4
2. Linear: First order ODEs of the formy 0 a(t)y b(t).(1)These can be solved by multiplying by the integrating factorRµ(t) ea(t) dtNotice that µ0 (t) a(t)µ(t) and so multiplying (1) by µ(t) we obtainµ(t)(y 0 a(t)y) µ(t)b(t)µ(t)y 0 µ0 (t)y µ(t)b(t)d(µ(t)y) µ(t)b(t).dtThus,1y µ(t) Z µ(t)b(t) dt3. Exact: First order ODEsM (x, y) N (x, y)y 0 0(2)that can be written asd(Ψ(x, y(x))) 0(3)dxfor some differentiable function Ψ(x, y). If it happens that M, N, M, N are all y xcontinuous in some rectangle then such function Ψ exists if and only if M N . y xIf an equation is exact, then the solutions are given byΨ(x, y) cwhere c is a constant. To find Ψ(x, y) we notice that for (2) to be equal to (3) werequire Ψ Ψ M, and N. x y5
ThereforeZΨ(x, y) M (x, y) dx C(y)for some function C(y), depending only on y, and that we determine from the Ψ N.condition y Some ODEs may not fall directly into any of the groups above, but after a modificationor substitution they can be solved by the same methods.1. Bernoulli equations: ODEs of the fromy 0 a(t)y b(t)y n .These equations can be transform into linear equations by using the substitutionu y 1 n .2. Integrating factors: Assume thatM (x, y) N (x, y)y 0 0(4)satisfies one of the following conditions:– The quotient M y N x P (x)Nis a function depending only on the variable x.– The quotient N M x y Q(y)Mis a function depending only on the variable y.ThenR (4) can be transformed into an exact equationR by multiplying by µ(x) exp( P (x) dx) in the first case, or by µ(y) exp( Q(y) dy) in the second case. ODEs are used to model real-life situations, some common applications are:1. Free falling objects: Let v(t) represent the velocity of an object at the time t.If the object falls freely (assuming there is no air resistance) thenmdvdv m(9.8) or 9.8,dtdt6
where 9.8m/s2 is the gravity constant, t is measured in seconds (s), and v(t) inmeters per second (m/s). If there is air resistance proportional to the velocity,say γv(t) for some constant γ, then we obtain the ODEdvγ 9.8 v,dtmwhich is separable. If instead, the object is thrown upwards then the gravity willact as a force of resistance, and under the assumption of an air resistance of theform γv(t), the associated ODE isγdv 9.8 v.dtm2. Exponential growth and decay: P (t) represents the size of some populationat the time t. If we assume that the rate of growth of P (t) is proportional to P (t)then we obtain the separable ODEdP kP.dt3. Predator/ Prey systems: Again P (t) represents the size of some populationat the time t and the rate of growth of P (t) is proportional to P (t). We alsoassume that due to the presence of some predator there are d deaths per unit oftime (here d is a constant). Then we obtain the ODEdP kP d,dtwhich is again separable.4. Newton’s law of cooling: Suppose that an object is initially in a room withtemperature T0 , and after a while is taken to the outside where the ambienttemperature is Ta . Newton’s Law of Cooling says that the rate of change of thetemperature of the object as the time passes is proportional to the differencebetween the ambient temperature and the temperature of the object, as an IVPthis can be written as:dT k(T Ta ), T (0) T0 ,dtfor some constant k 0. This ODE is separable.7
5. Tank model: Suppose that a tank contains V0 gallons of brine with y0 poundsof salt in it. Suppose that brine with a concentration of sin pounds of salt/gallonis poured into the tank at a rate of νin gallons/minute, and that the well mixedliquid comes out of the tank at a rate of νout gallons/minute. If y(t) representsthe amount of salt (in pounds) in the tank at the time t (in minutes), thendy (rate in) (rate out)dtSince rate in (out) number of pounds of salt poured in (coming out) per minute,then assuming that the mixture is homogenous we obtainrate in sin νiny(t) νout .rate out V (t)Notice that the volume of liquid in the tank at the time t is given byV (t) V0 νin t νout t.To determine the amount of salt in the tank at the time t we have to solve theIVPdyy sin νin νout , y(0) y0 .dtV (t)This is a linear ODE. A second order linear ODE is a differential equation of the forma(t)y 00 b(t)y 0 c(t)y d(t)(*) The associated homogeneous equation isa(t)y 00 b(t)y 0 c(t)y 0(?) Two solutions y1 (t) and y2 (t) for (?) are said to form a fundamental set of solutions ifthey are linearly independent, i.e., if the WronskianW (y1 , y2 )(t) y1 (t) y2 (t) y1 (t)y20 (t) y10 (t)y2 (t) 6 0.y10 (t) y20 (t)8
If y1 and y2 form a fundamental set of solutions for (?), then every other solution is ofthe formYH (t) c1 y1 (t) c2 y2 (t)for some constants c1 and c2 . We refer to YH (t) as the general solution to the homogeneous equation (?). Suppose that you know a particular solution YP (t) for (*), then if y(t) is any othersolution one can easily show that y(t) YP (t) is a solution for (?) and soy(t) YH (t) YP (t)Thus, the general solution for (*) consists of the general solution for the associatedhomogeneous plus a particular solution. When the coefficients of (?) are constant then it is easy to find a fundamental set ofsolutions. Consider the equationay 00 by 0 cy 0.(z)When looking for solutions of the form y ert one finds out that r must satisfy thecharacteristic equationar2 br c 0.Of course, the solutions to the characteristic equation vary depending on the values ofthe constants a, b, c and we have the following cases:1. Two real solutions: This happens when b2 4ac 0, in which case the solutionsare b b2 4ac b b2 4ac, and r2 .r1 2a2aIn this case the solutions y1 er1 t and y2 er2 t form a fundamental set ofsolutions and therefore the general solution for (z) isy(t) c1 er1 t c2 er2 t9
2. Two complex solutions: This happens when b2 4ac 0, in which case thesolutions are b4ac b2 b4ac b2r1 i, and r2 i.2a2a2a2aIn this case the solutions y1 er1 t and y2 er2 t are complex, which may ormay not be appropriate depending on what the solutions of the equation shouldrepresent. Fortunately, one is able to prove that if y(t) A(t) iB(t) is a solutionfor (z) then so are the real part A(t) and the imaginary part B(t). Using themore convenient notation 4ac b2 band ω ,r1 λ iω, where λ 2a2aone obtains the solutiony(t) eλt iω t eλt eiωt eλt (cos ωt i sin ωt).The real and imaginary parts of y(t) are the solutionsy1 (t) eλt cos ωt and y2 (t) eλt sin ωt.After computing W (y1 , y1 ) one sees that y1 and y2 form a fundamental set ofsolutions for (z) and therefore the general solution isy(t) c1 eλt cos ωt c2 eλt sin ωt3. Repeated roots: This happens when b2 4ac, in which case we obtain only onesolution to the characteristic equationr λ b.2aIn this case one finds out (using the technique of reduction of order) that afundamental set of solutions is given by y1 (t) eλt and y2 (t) teλt and so thegeneral solution for (z) isy(t) c1 eλt c2 teλt10
We are now interested in finding particular solutions to second order ODEs whoseassociated homogeneous equation has constant coefficients, i.e., equations of the formay 00 by 0 cy d(t).(zz)1. Undetermined Coefficients: The idea is to look for a particular solution YP (t)that looks like d(t). We follow the following table:d(t)Pn (t)Pn (t)eαtPn (t)eαt sin βtPn (t)eαt cos βtYP (t)ts Qn (t)ts Qn (t)eαtts eαt [Qn (t) cos βt Rn (t) sin βt]ts eαt [Qn (t) cos βt Rn (t) sin βt]wherePn (t) An tn · · · A1 t A0Qn (t) Bn tn · · · B1 t B0Rn (t) Cn tn · · · C1 t C0 .2. Variation of Parameters: If it happens that the function d(t) is nothing likethe functions in the table above, then we can try to look for a particular solutionof the formYP (t) u1 (t)y1 (t) u2 (t)y2 (t).After one extra assumption one is able to deduce formulas for u1 and u2 in termsof y1 , y2 , and d. These formulas are:Zu1 (t) y2 (t)d(t)dtaW (y1 , y2 )(t)Zandu2 (t) y1 (t)d(t)dtaW (y1 , y2 )(t)NOTE: These integrals may be very difficult to compute. Suppose that a spring of length in vertical position is stretched to a length L,i.e., L units beyond its natural length, by a mass m attached to its end. Then Hooke’sLaw states that there is a constant k (independent of L and m) such thatmg kL.11
Suppose that the spring is stretched further a distance u(t), then one obtains the secondorder ODE:Xmu00 (t) forces acting on m (weight) (force due to the spring) (damping force) (total external force) mg k(L u(t)) γu0 (t) F (t) (mg kL) ku(t) γu0 (t) F (t) ku(t) γu0 (t) F (t).Here we are using the fact that mg kL 0, and we are assuming that for small u(t)the damping force (due to air resistance or viscosity) is proportional to the velocityu0 (t). The constant γ is called damping constant. Then we obtain a second orderODE with constant coefficients, which will have unique solution once we specify initialposition and velocity, i.e., we have the following IVP:mu00 (t) γu0 (t) ku(t) F (t), u(0) u0 , u0 (0) u00 Recall that if a function f (t) is continuous in every interval of the form [0, b] for b 0,then one can define the infinite integralZ Z bf (t) dt : limf (t) dt.b 00If such limit exists (it is a number) then we say that the integral converges, otherwise(the limit is or does not exist) we say that the integral diverges. Assume that a function f (t) is continuous in every interval of the form [0, b] for b 0,and assume further that there are constants K, M, α such that f (t) Keαt , for all t M.Then the integralZL{f (t)} : F (s) e st f (t) dt0converges for all s α. We refer to the function L{f (t)} as the Laplace Transformof f (t).12
It follows from the additive properties of the integral that the Laplace transform islinear, i.e.,L{af (t) bg(t)} aL{f (t)} bL{g(t)},for any constants a, and b. If L{f (t)} F (s) then we write L 1 {F (s)} f (t). L 1 is also linear and is calledthe Inverse Laplace Transform. For a real constant c we define the step function(1 if t cuc (t) 0 otherwise. The following list of common Laplace transforms will be provided to you in the exam:13
f (t) L 1 {F (s)}F (s) L{f (t)}11, s 0seat1, s as an!tn , n positive integersn 1, s 0s, s 0s 2 a2cos atsin ats2cosh atsinh ata, s 0 a2s2s, s a a2s2a, s a a2eat cos bts a, s a(s a)2 b2eat sin btb, s a(s a)2 b2tn eat , n positive integern!, s a(s a)n 1uc (t)e cs, s 0suc (t)f (t c)e cs F (s)ect f (t)F (s c)14
Laplace transforms provide a valuable tool when solving IVP with non homogeneouslinear ODEs involving piecewise defined functions. For second order linear ODEs theadvantage becomes evident from the relationsL{y 0 (t)} sL{y(t)} y(0)L{y 00 (t)} s2 L{y(t)} sy(0) y 0 (0) The algorithm to solve an IVP using Laplace transform is very simple:1. Apply Laplace transform to both sides of the equation.2. Use the relations for L{y 00 } and L{y 0 } boxed above.3. Use the given initial conditions y(0) y0 and y 0 (0) y00 .4. Solve for L{y}.5. Apply L 1 to obtain y. This step is critical since computing L 1 may be challenging, the idea is to rewrite the function of s as a linear combination of functionsappearing in the big table of Laplace transforms. A system of first order linear ODEs is an equation of the form:x0 (t) P(t)x(t) g(t),where x1 (t) x2 (t) x(t) . , . xn (t) p1,1 (t) p1,2 (t) p2,1 (t) p2,2 (t) P(t) . .pn,1 (t) pn,2 (t)······.··· p1,n (t)p2,n (t) . ,. pn,n (t) g1 (t) g2 (t) g(t) . . . gn (t) Though the theory we will describe in what follows applies to matrices of any size, wewill focus on the 2 2 and 3 3 cases. We will also assume that the matrix P(t) isconstant and we will denote it by A. A matrix A has an inverse if and only if det(A) 6 0. a b Assume that the matrix A is invertible. Then its inverse if given byc dA 11 ad bc15 d b c a
Recall that the (column) vectors X (1) , X (2) , . . . , X (n) are linearly independent if andonly if det X (1) X (2) · · · X (n) 6 0. If X (1) (t), X (2) (t), . . . , X (n) (t) are linearly independent solutions for the homogeneoussystemx0 Ax,then the general solution is of the formx(t) c1 X (1) (t) c2 X (2) (t) · · · cn X (n) (t)In this case the matrix (1) Ψ(t) XX (2) · · · X (n) is called a fundamental matrix for the system. Thus the general solution can bewritten as c1 c2 x(t) Ψ(t)c, where c . . cn To solve the IVPx0 Ax, x(t0 ) x0 ,we look for the the vector c such thatx(t0 ) Ψ(t0 )c x0 .Since the matrix Ψ(t) is invertible thenc (Ψ(t0 )) 1 x016
The eigenvalues of a square matrix are the values λ satisfying the equation:det(A λI) 0 If det(A λI) 0 then the system of equations corresponding to the matrix A λIadmits more than one solution and so it is possible to find a nonzero vector ξ such that(A λI)ξ 0Such vector is called an eigenvector associated to the eigenvalue λ. If A is square matrix whose entries are complex numbers, then A can be written asA TJT 1for some invertible matrix T. The matrix J is upper triangular and carries all therelevant information from A. J is called the canonical Jordan from of A. In the casewhen A is size 2 2, we have the following possibilities:1. A has two different eigenvalues λ1 and λ2 . In this case λ1 0J , and T T (1) T (2) ,0 λ2 where T (1) is an eigenvector for the eigenvalue λ1 , and T (2) is an eigenvector forthe eigenvalue λ2 .2. A has only one eigenvalue λ, but it has two linearly independent eigenvectors T (1)and T (2) . In this case λ 0J , and T T (1) T (2) .0 λ 3. A has only one eigenvalue λcase λJ 0and one eigenvector T (1) (up to multiples). In this 1, and T T (1) T (2) ,λ where the vector T (2) is solution to(A λI)T (2) T (1) .17
To solve an homogeneous system of linear ODEs with constant coefficientsx Axwe follow the following steps:1. Compute the canonical Jordan form of the matrix A.2. Use the substitution x Ty. In this case we get x0 Ty0 and therefore weobtain the new systemTy0 ATy.3. Multiplying both sides by T 1 the system becomesy0 Jy.Reading out the two equations we obtain two very simple first order linear ODEs,solving first the equation for y2 (t) and then the equation for y1 (t) we obtain y.4. We find x from our substitution x Ty. The output of the algorithm above will depend on the shape of J. λ1 01. If J and T T (1) T (2) then0 λ2x(t) c1 T (1) eλ1 t c2 T (2) eλ2 tThe same answer if there is only one eigenvalue (i.e., λ λ1 λ2 ), but twolinearly independent eigenvectors T (1) and T (2) . λ 12. If J , T (1) is an eigenvector and T (2) satisfies0 λ(A λI)T (2) T (1) ,then the solution isx(t) c1 T (1) eλt c2 T (1) teλt T (2) eλt It could happen that the entries of a matrix A are all real numbers, but A has complexeigenvalues. In this case, we can take the real and imaginary parts of the generalsolution to obtain a fundamental set of real solutions. We show this with an example:18
Example 1. Find the general solution to the system 0 10x x. 1 0First, we need to find the eigenvalues: λ 1det λ2 1 0 λ i. 1 λEigenvector for λ1 i: i 1ξ10 i 1 i 1 , 1 iξ20 1 i0 0Thus, iξ1 ξ2 0 and therefore the solutions are of the form ξ1ξ11 ξ1.ξ2iξ1i 1(1)We choose the eigenvector T . Similarly for λ2 i we obtain the eigenvectori 1T (2) . i i 01 1The Jordan form is J , and T .0 ii iThe general solution is 1 it1x(t) c1e c2e iti i 11 c1(cos t i sin t) c2(cos t i sin t)i i c1 cos t ic1 sin t c2 cos t ic2 sin t c1 sin t ic1 cos t c2 sin t ic2 cos t c1 cos t c2 cos tc1 sin t c2 sin t i c1 sin t c2 sin tc1 cos t c2 cos t cos tsin t (c1 c2 ) i(c1 c2 ). sin tcos t19
Thus, the general solution can be written as cos tsin tx(t) A B sin tcos tfor constants A and B. As in the case of second order ODEs, the general solution to the system of linear ODEsx0 (t) Ax(t) g(t)can be written as x(t) xH (t) xP (t), where xH is the general solution to theassociated homogeneous system and xP is a particular solution. To solve a non-homogeneous system of linear ODEs of the formx0 (t) Ax(t) g(t)we follow the following steps:1. Compute the canonical Jordan form of the matrix A (this includes J, T, andT 1 ).2. Use the substitution x Ty. In this case we get x0 Ty0 and therefore weobtain the new systemTy0 ATy g(t).3. Multiplying both sides by T 1 the system becomesy0 Jy T 1 g(t).Reading out the two equations we obtain two first order linear ODEs, solving firstthe equation for y2 (t) and then the equation for y1 (t) we obtain y.4. We find x from our substitution x Ty. Even though for 2 2 systems you can just solve the equation by finding the Jordanform of the matrix A and using the substitution x Ty as explained above, forsystems of more variables it is sometimes quicker to find a particular solution by usingeither undetermined coefficients or variation of parameters:20
1. Undetermined Coefficients: The idea is to look for a particular solution xP (t)that looks like g(t). We follow the table:xP (t)ts Qn (t)g(t)Pn (t)Pn (t)eαtPn (t)eαt sin βtPn (t)eαt cos βtts Qn (t)eαtts eαt [Qn (t) cos βt Rn (t) sin βt]ts eαt [Qn (t) cos βt Rn (t) sin βt]wherePn (t) an tn · · · a1 t a0Qn (t) bn tn · · · b1 t b0Rn (t) cn tn · · · c1 t c0 .NOTE: Here an , . . . , a0 , bn , . . . , b0 , cn , . . . , c0 are constant vectors.2. Variation of Parameters: We know that the solution to the associated homogeneous system x0 Ax can be written asxH (t) Ψ(t)c.The idea is to look for a particular solution of the formxP (t) Ψ(t)u(t)for some vector u1 (t) u2 (t) u(t) . . . un (t)Computing derivatives we obtainx0P (t) Ψ0 (t)u(t) Ψ(t)u0 (t).Replacing into the equation we getx0P (t) Ψ0 (t)u(t) Ψ(t)u0 (t) AΨ(t)u(t) g(t).The second equality can be rewritten as(Ψ0 (t)u(t) AΨ(t)u(t)) Ψ(t)u0 (t) g(t) {z} 021
The first term vanishes because Ψ0 (t)c AΨ(t)c for every vector c, includingu(t). Therefore we must haveΨ(t)u0 (t) g(t), or u0 (t) (Ψ(t)) 1 g(t).ThusZtu(t) (Ψ(t)) 1 g(t) dtt0As for the case of second order ODEs with constant coefficients, these integralsmay be very hard to compute. As in the case of second order ODEs with constant coefficients, it is possible to useLaplace transform to solve IVP of the formx0 (t) Ax(t) g(t), x(0) x0 .Define X1 (s)L{x1 (t)}x1 (t) x2 (t) L{x2 (t)} X2 (s) L(x(t)) L . . X(s). . . . xn (t)L{xn (t)}Xn (s)Then it follows thatL{x0 (t)} sL{x(t)} x(0) sX(s) x0Applying Laplace Transform to the system we obtainsX(s) x0 AX(s) G(s),where G(s) L{g(t)}. Solving for X(s) the equation becomesX(s) (sI A) 1 (G(s) x0 ).Therefore, the solution is given by x(t) L 1 (sI A) 1 (G(s) x0 )To compute the Laplace inverse one needs to compute (sI A) 1 (G(s) x0 ) as amatrix of s and as usual apply partial fractions to write each entry of the matrix ascombinations of the functions appearing in the table for Laplace transforms.22
MATH 4B{Di erential Equations, Fall 2016 Final Exam Study Guide GENERAL INFORMATION AND FINAL EXAM RULES The exam will have a duration of 3 hours. No extra time will be given. Failing to submit your solutions within 3 hours will result in your exam not being graded. The Final Exam is comprehensive. The sections are 1.1{1.3, 2.1{2.6, 3.1{3.7, 6 .
