Math 209: Important Formulas

These notes give a brief summary of the major techniques of the class, and an example for each. The descriptions are kept short so that this can be a useful, quick reference. 1. Executive Summary We record below the types of equations we can solve. In the next sections we give more details, including conditions on the functions and, if possible, explicit solutions. This section is meant to be a quick list. • Linear, constant coefficient difference equations: an+1 = c1an+c2an−1+c3an−2+ · · ·+ ckan−k+1; for example, an+1 = 3an + 4an−1 − 2an−2. See §2. • Integrating factors: dy/dt+ p(t)y(t) = g(t). See §3.1. • Separable equations: M(x) +N(y)dy/dx = 0. See §3.2. • Exact equations: M(x, y) +N(x, y)dy/dx = 0 with ∂M/∂y = ∂N/∂x. See §3.3. • Second order linear constant coefficient homogenous equations: dy/dt + ady/dt+ by = 0. See §4.1. • Method of Undermined Coefficients: ay′′ + by′ + cy = g(t) with g(t) = ePn(t) (with Pn(t) a polynomial of degree n in t) or g(t) = e αt cos(βt) or g(t) = e sin(βt). See §4.2. • Variation of Parameters: Let p, q, g continuous functions and consider y′′(t) + p(t)y′(t) + q(t)y = g(t) with known solutions to the homogenous equation. See §4.3. • Series expansions: p(x)y′′(x)+q(x)y′(x)+r(x)y(x) = 0 with p(x0) 6= 0 and guessing y(x) = ∑∞ n=0 an(X − x0). See §5. • Linear systems: − →x ′(t) = A− →x (t) +− →g (t) with − →x (0) = − →x 0. See §6. Date: April 28, 2009. 1 2 INSTRUCTOR: STEVEN MILLER 2. Difference Equations 2.1. Linear, constant coefficient difference equations. Statement: Let k be a fixed integer and c1, . . . , ck given real numbers. Then the general solution of the difference equation an+1 = c1an + c2an−1 + c3an−2 + · · ·+ ckan−k+1 is an = γ1r n 1 + · · ·+ γkr k if the characteristic polynomial r − c1r − c2r − · · · − ck = 0 has k distinct roots. Here the γ1, . . . , γk are any k real numbers; if initial conditions are given, these conditions determine these γi’s. Example: Consider the equation an+1 = 5an − 6an−1. In this case k = 2 and we find the characteristic polynomial is r − 5r + 6 = (r − 2)(r − 3), which clearly has roots r1 = 2 and r2 = 3. Thus the general solution is an = γ12 n + γ23 . If we are given a0 = 1 and a1 = 2, this leads to the system of equations 1 = γ1 + γ2 and 2 = γ1 · 2 + γ2 · 3, which has the solution γ1 = 1 and γ2 = 0. Applications: Population growth (such as the Fibonacci equation), why double-plus-one is a bad strategy in roulette. 3. First Order Differential Equations 3.1. Integrating factors. Statement: For a differential equation of the form y′(t) + p(t)y(t) = g(t), the general solution is y(t) = 1 μ(t) [∫ μ(t)g(t)dt+ C ] , where μ(t) = exp (∫ p(t)dt ) and C is a free constant (if an initial condition is given, then C can be determined uniquely). Example: Consider the equation y′(t)− 2ty(t) = exp (t + t). Then μ(t) = exp (∫ −2tdt ) = exp(−t), MATH 209: IMPORTANT FORMULAS 3