Review Sheet for Ordinary Differential Equations | MA 26600, Study notes of Differential Equations

Download Review Sheet for Ordinary Differential Equations | MA 26600 and more Study notes Differential Equations in PDF only on Docsity! A Brief Review Chapter 3: Second Order Linear Equations Chapter 4: Higher Order Linear Equations If you have time, for each iterm I list below, you should do at least two practice problems. 1 Chapter 3: Second Order Linear Equations 1. Knowing how to determine the longest interval in which the solution to the initial value problem y ′′ + p(t)y ′+ q(t)y = g(t), y(t0) = y0, y ′(t0)= y0 ′ exists, without solving the initial value problem. (Answer: the longest interval containing t0 in which p(t), q(t) and g(t) are continuous. If the coefficients of y ′′ in the given equation is not one, you need first divide the equation by that coefficients. ) Practice Problems: Section 3.2 #8, 9 2. Homogeneous Equations with constant coefficients a y ′′+ b y ′+ c y = 0 (1) First solve the characteristic equation, which is a r2 + b r + c =0. (2) Using quadratic formula, the roots are given by r1 = − b + b2− 4 a c √ 2 a , r2 = − b− b2− 4 a c √ 2 a There might be three cases: i. r1, r2 are real and different roots(when b2− 4ac> 0): then the general solution of (1) is given by y(t) = c1 er1t + c2er2t (Practice Problems: section 3.1 #1,4, 10) ii. r1,r2 are complex conjugates(when b 2 − 4 ac < 0): suppose r1 = λ + iµ, r2 = λ − iµ, the two real-valued solutions are (which are real part and imaginary part of e(λ+iµ)t ) u(t)= eλtcos(µt), v(t)= eλtsin(µt). The the general solution is given by y(t)= c1 e λt cos(µt)+ c2 e λt sin(µt) (Practice Problems: Section 3.4(Section 3.3 of 9th edition) #10, 17, 18) iii. real repeated roots(when b2− 4ac= 0): r1 = r2 =− b 2a = r, in this case, the general solution is given by y(t)= c1 e rt + c2 t e rt (Practice Problems: Section 3.5(3.4 in 9th edition) #3, 12) 3. Nonhomogeneous Equations General form: y ′′+ p(t)y ′+ q(t)y = g(t) (3) 1 With constant coefficients: a y ′′+ b y ′+ c y = g(t), a 0 (4) The general solution of (3) (or (4)) can be written as y(t)= c1y1(t)+ c2y2(t)+ Y (t), where c1y1(t)+ c2y2(t) is the general solution of the corresponding homogeneous equation, and Y (t) is a particular solution of the nonhomogeneous equation. There are two methods to find a partic- ular solution Y (t) i. Undetermined Coefficients (IMPORTANT) This method is applicable to equation (4) when g(t) is an exponential function, sine, cosine function, polynomial function or their product or sum. The key step is to choose the right form of a particular solutino Y (t). Table 3.6.1 on page 181 of 8th edition textbook(Is is Table 3.5.1 on 9th edition) tell you how to choose the right form. (Please do go over all the homework of this section) ii. Variation of Parameters This method is more powerful, because it is applicable to the general equation (3). Suppose you can find two solutions y1(t) and y2(t) of the corresponding homogeneous equation with nonzero Wronskian W [y1, y2](t)  0, then a particular solution Y (t) of the nonhomogeneous equation is given in following form Y (t)= u1(t)y1(t) +u2(t) y2(t), (5) where u1(t)=− ∫ y2(t)g(t) W [y1, y2](t) dt, u2(t)= ∫ y1(t)g(t) W [y1, y2](t) dt (6) (Remember this formula, and do at least one practice problem. You won’t have enough time to repeat the full procedure of variation of parameters in exam) Practice Problems: Section 3.7 (3.6 in 9th edition) #7, 10, 13 4. Application: Spring-Mass system the governing equation of a spring-mass system m u′′(t)+ γ u′(t)+ k u(t)= F (t), (7) where m is the mass, γ is the damping coefficient, and k is the spring constant. Pay special atten- tion to their units when you build the eqaution. You should know how to find the amplitude, period (or quasi period) of the solution, and know the differences between γ = 0 and γ  0. You needn’t remember any formula except the governing equation (7). Practice Problems: Section 3.8 (3.7 in 9th edition) #5, 9 2 Chapter 4: Higher Order Linear Equations This chapter is an extension of Chapter 3, which means most of the results of second order linear equa- tions can be directly extended to higher order linear equations. Following are the differences 1. For a higher order linear equation L[y] = a0y (n) + a1 y (n−1) + + an−1 y ′+ an y = 0 (8) its characteristic equation is a0 r n + a1r n−1 + + an−1 r + an =0 (9) 2 Section 2