SECOND ORDER ODE S. 1. A second order differential equation is an equation of the form. F (x, y, y, y ) = 0.

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1 SECOND ORDER ODE S 1. A second der differential equation is an equation of the fm F (x, y, y, y ) = 0. A solution of the differential equation is a function y = y(x) that satisfies the equation. A differential equation has infinitely many solutions. They usually involve two undetermined constants. 2. An initial value problem consists of a differential equation and an initial conditions y(x 0 ) = y 0, y (x 0 ) = y 0. It has a unique solution, which has to be a function (provided F doesn t have problems at that point). 3. A boundary problem consists of a differential equation and conditions y(x 0 ) = y 0, y(x 1 ) = y 1 ; y (x 0 ) = y 0, y(x 1 ) = y 1 ; y(x 0 ) = y 0, y (x 1 ) = y 1; y (x 0 ) = y 0, y (x 1 ) = y 1. A. Homogeneous linear second der ODEs with constant coefficients mx + bx + kx = 0, m, b, k constants, m 0 1. If the equation describes a physical spring dashpot system, the coefficients m, b, k are non-negative. 2. An initial value problem consists of a differential equation and an initial condition x(t 0 ) = A, x (t 0 ) = B. How many solutions does it have? 3. The general solution is of the fm x(t) = c 1 x 1 (t) + c 2 x 2 (t), where x 1 and x 2 are two linearly independent solutions (none of them can be written as a constant multiple of the other). 4. The characteristic polynomial of this equations is p(s) = ms 2 + bs + k. 5. The exponential solutions of this equation are c 1 e r1t and c 2 e r2t, where r 1, r 2 are the roots (real complex) of the characteristic polynomial and c 1, c 2 are arbitrary constants. If r 1 = r 2 = r, there is only one family of exponential solutions, namely ce rt. How to solve: 1. Write down the characteristic equation ms 2 + bs + k = Compute its discriminant = b 2 4mk. 3. There are three possible situations: 1

2 overdamped: If > 0, the quadratic equation has two distinct real solutions, r 1 and r 2. Find them. (You might need to use the quadratic fmula.) The general solution of the differential equation is x = C 1 e r1t + C 2 e r2t. critically damped: If = 0, the quadratic equation has only one real root r. Find it. The general solution of the differential equation is x = C 1 e rt + C 2 te rt. underdamped: If < 0, the quadratic equation does not have any real roots, it has two complex conjugate roots r 1,2 = α ± iβ, where α = b 2m and β = 2m. The general solution of the differential equation is x = C 1 e αt cos(βt) + C 2 e αt sin(βt) x = Ae αt cos(βt φ). In the second expression, A is any positive real number and φ any number in [0, 2π). 4. If it is an initial value problem a boundary problem, plug in the given values and solve f C 1 and C 2 ( f A and φ). Don t fget to take the derivative of x in the case of an initial value problem (chain rule!). B. Nonhomogeneous second der linear ODEs mx + bx + kx = f(t). We solve these by finding a particular solution x p of the given ODE and the solution x h to the cresponding homogeneous equation mx + bx + kx = 0. The general solution is given by x = x p + x h. Note: x h contains all the undetermined constants (denotes an infinite family of functions), while x p is one particular function. We have two ways of finding x p, namely ERF and undetermined coefficients. They are both explained below f linear equations of arbitrary der. Just apply them f your second der ODE. A. Linear operats LINEAR ODEs AND LINEAR OPERATORS Notation D stands f the derivative of a function. F instance, if x is a function of t, Dx = x = dx/dt. If f is a function of x, then Df = f = df/dx. D n stands f taking the n-th derivative. F instance D 2 x = D(Dx) = D(x ) = x. I is the identity operat that leaves functions alone, so I(f) = f. In general p(d) = a n D n a 1 D + a 0 I applied to a function y means taking a linear combination of y and its first n derivatives, namely p(d)y = (a n D n a 1 D + a 0 I)y = a n y (n) a 1 y + a 0 y Writing p(d)y = f(t) provides an handy shthand f the linear ODE a n y (n) a 1 y + a 0 y = f(t) 2

3 Properties Linearity p(d)(af + bg) = ap(d)f + bp(d)g Exponential shift fmula p(d)(e rt u) = e rt p(d + ri)u Don t fget that (D + ri) 2 = D 2 + 2rD + r 2 I (and so fth). B. Homogeneous linear ODEs with constant coefficients This equation can be re-written as with a n y (n) a 1 y + a 0 y = 0 p(d)y = 0 p(d) = a n D n a 1 D + a 0 I 1. Write down the characteristic polynomial p(s) = a n s n a 1 s + a Find its n roots r 1,..., r n (counted with multiplicity) by solving a n r n a 0 = The general solution is of the fm x = c 1 x c n x n where each x j cresponds to the root r j as follows. F each simple real root r j we obtain an exponential solution x j = e rjt. F each pari of simple complex conjugate roots r j,j+1 = a ± ib we get the pair of solutionsx j = e at cos bt and x j+1 = e at sin bt A repeated real root r j with multiplicity m, should have m cresponding solutions. Construct x j as above. The rest are tx j, t 2 x j,..., t m 1 x j. F repeated pairs of complex roots the same principle applies, except now you have m pairs of cresponding solutions. C. Non-homogeneous linear ODEs This equation can be re-written as with a n y (n) a 1 y + a 0 y = 0 p(d)y = 0 p(d) = a n D n a 1 D + a 0 I We solve these by finding a particular solution x p of the given ODE and the solution x h to the cresponding homogeneous equation mx + bx + kx = 0. The general solution is given by x = x p + x h. Note: x h contains all the undetermined constants (denotes an infinite family of functions), while x p is one particular function. We have two ways of finding x p, detailed below. 3

4 Exponential response fmulas: can be applied to an equation with exponential input where r is any complex number. p(d)x = Ae kt, 1. Write down the characteristic polynomial p(s). 2. Compute p(k). If it is nonzero, then x p = Aekt p(k) is a solution. (ERF) 3. If p(k) = 0, compute p (k). If this in nonzero, then x p = Atekt p (k) is a solution. (ERF ) 4. Keep going. The idea is to compute p(k), p (k), p (k),... until you find the first nonzero guy amongst them. Say that happens to be p (m) (k). Then is a solution. x p = Atm e kt p (m) (k) Undetermined coefficients method: the idea is to look f a solution of the same general fm as the function f(t) on the RHS of our equation. If f(t) is of the fm ae rt a cos ωt a sin ωt try Ae rt A cos ωt + B sin ωt A cos ωt + B sin ωt a n t n a 0 A n t n A 0 (a n t n a 0 )e rt (A n t n A 0 )e rt (a n t n a 0 ) cos ωt (A n t n A 0 )(B 1 cos ωt + B 2 sin ωt) (a n t n a 0 ) sin ωt (A n t n A 0 )(B 1 cos ωt + B 2 sin ωt) (a n t n a 0 )e rt cos ωt (A n t n A 0 )e rt (B 1 cos ωt + B 2 sin ωt) (a n t n a 0 )e rt sin ωt (A n t n A 0 )e rt (B 1 cos ωt + B 2 sin ωt) If f(t) = f 1 (t) + f 2 (t), look f solutions x j (t) cresponding to f j (t), j = 1, 2, and then add them up to get x p = x 1 + x 2. G. Physical notions complex gain amplitude gain 4

5 phase lag/shift time lag amplitude resonance practical resonance frequency pseudofrequency 5