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HANDOUT I 1 Linear first-order equations A first-order linear ODE is an equation of the form a1 (t)y 0 + a0 (t)y = b(t), (1.1) a1 (t), a0 (t) are called coefficient functions, b(t) is the source term. A linear ODE is called homogeneous if the source term is 0. Every homogeneous first-order ODE can be written as a0 (t) y 0 = k(t)y, k(t) := − (1.2) a1 (t) The general solution to (1.2) is R y = Ce k(t)dt = Ce − R a0 (t) dt a1 (t) (1.3) The general solution to inhomogeneous equation (1.1) takes the form: y = yp (t) + Cyh (t), (1.4) where yp (t) is a particular solution to (1.1) and yh (t) is a particular nontrivial solution to corresponding homogeneous equation (1.2). In order to find (1.4) we use the method of variation of constants: substitute y = v(t)yh (t) in (1.1) and solve for v(t). Example. y 0 + 4y = 2e−4t sin(2t) (1.5) The corresponding homogeneous equation: y 0 = −4y (1.6) By formula (1.3) the general solution to (1.6) is y = Ce−4t (1.7) Now find the general solution to (1.5) in the form: y = v(t)e−4t (1.8) Substitute (1.8) into (1.5) and get v 0 = 2 sin(2t) Thus v(t) = − cos(2t) + C and therefore the general solution to (1.5) is y(t) = −e−4t cos(2t) + Ce−4t 1 (1.9) 2 Second-order linear equations A second-order linear ODE is an equation of the form a2 (t)y 00 + a1 (t)y 0 + a0 (t)y = b(t) (2.1) The corresponding homogeneous equation: a2 (t)y 00 + a1 (t)y 0 + a0 (t)y = 0 (2.2) The general solution to (2.2) is yg.h = C1 y1 (t) + C2 y2 (t), (2.3) where y1 , y2 are two linearly independent solutions to (2.2). The general solution to inhomogeneous equation (2.1): yg.i = C1 y1 (t) + C2 y2 (t) + yp (t), (2.4) where yp (t) is a particular solution to (2.1). I. Homogeneous equations with constant coefficients. a2 y 00 + a1 y 0 + a0 y = 0 (2.5) To solve (2.5) consider the characteristic equation: a2 s2 + a1 s + a0 = 0 (2.6) The general solution to (2.5) is 1. y = C1 er1 t + C2 er2 t , if the characteristic roots r1 and r2 are real and distinct. 2. y = ept (C1 cos(qt) + C2 sin(qt)), if the characteristic roots are complex conjugate p ± qi. 3. y = er1 t (C1 + C2 t), if there is a double characteristic root r1 . II. Inhomogeneous equation (2.1). If one knows y1 (t) and y2 (t), two linearly independent solutions to (2.2), then it is possible to find the general solution to (2.1) by variation of constants: substitute y = v1 (t)y1 (t)+v2 (t)y2 (t) in (2.1) and solve the following system for v10 (t) and v20 (t): v10 y1 + v20 y2 = 0 v10 y10 + v20 y20 = b(t)/a2 (t) Example. Solve the equation y 00 − 2y 0 + y = 2 ex x (2.7) The corresponding homogeneous equation is y 00 − 2y 0 + y = 0 (2.8) The characteristic equation: s2 − 2s + 1 = 0, s1,2 = 1 (2.9) Find the general solution to (2.7) in the form yg.i = v1 (x)ex + v2 (x)xex (2.10) The functions v10 (x) and v20 (x) must satisfy the system: v10 ex + v20 xex = 0 v10 ex + v20 ex (1 ex + x) = x From the above system one gets v10 (x) = −1, v20 (x) = 1 x Thus v1 (x) = −x + C1 , v2 (x) = ln |x| + C2 and yg.i = (C1 − x)ex + (ln |x| + C2 )xex = ex (A1 + A2 x + x ln |x|) (2.11) III. In order to solve the equation a2 y 00 + a1 y 0 + a0 y = b(t) (2.12) in some cases one can use the method of undetermined coefficients. Example. Solve the equation y 00 − 3y 0 + 2y = x cos x (2.13) The corresponding homogeneous equation is y 00 − 3y 0 + 2y = 0 (2.14) The characteristic equation: s2 − 3s + 2 = 0, s1 = 1, s2 = 2 (2.15) Therefore yg.h = C1 ex + C2 e2x 3 (2.16) The right-hand side of (2.13) is x cos x, a first degree polynomial multiplied by cos x, so we are looking for the particular solution to (2.13) in the form yp = (A1 x + B1 ) cos x + (A2 x + B2 ) sin x (2.17) yp0 = (A1 + B2 + A2 x) cos x + (A2 − B1 − A1 x) sin x (2.18) One has and yp00 = (2A2 − B1 − A1 x) cos x − (2A1 + B2 + A2 x) sin x (2.19) Substitute yp (x) and its derivatives in (2.13) and get the following system for A1 , A2 , B1 and B2 : −3A1 + 2A2 + B1 − 3B2 = 0 A1 − 3A2 = 1 −2A1 − 3A2 + 3B1 + B2 = 0 3A1 + A2 = 0 This implies A1 = 0.1, A2 = −0.3, B1 = −0.12, B2 = −0.34 Hence yp = C1 ex + C2 e2x + (0.1x − 0.12) cos x − (0.3x + 0.34) sin x (2.20) IV. Reduction of order. The reduction of order procedure is designed to find the general solution of inhomogeneous equation (2.1). It can be applied if a nontrivial solution y1 (t) of the associated homogeneous equation (2.2) is known. The procedure is as follows: substitute y(t) = v(t)y1 (t) in equation (2.1), where v(t) is a new dependent variable. When simplified, equation (2.1) becomes a first-order linear ODE with dependent variable w(t) := v 0 (t). 3 Homework 1. Solve first-order linear equations y 0 + y tan x = sec x (y = sin x + C cos x) y = x(y 0 − x cos x) (y = x(C + sin x)) (y = C ln2 x − ln x) (xy 0 − 1) ln x = 2y 4 2. Solve second-order linear equations with constant coefficients y 00 − 3y 0 + 2y = sin x (y = C1 ex + c2 e2x + 0.1 sin x + 0.3 cos x) y 00 − 4y 0 + 8y = e2x + sin 2x (y = e2x (C1 cos 2x + C2 sin 2x) + 0.25e2x + 0.1 cos 2x + 0.05 sin 2x) 3. Solve second-order linear equations with variable coefficients by reduction of order x2 y 00 ln x − xy 0 + y = 0, y1 = x (y = C1 x + C2 (ln x + 1)) xy 00 − (2x + 1)y 0 + (x + 1)y = 0 y1 = ex (y = ex (C1 x2 + C2 )) 4. Solve by variation of constants y 00 + 3y 0 + 2y = y 00 + y = ex 1 +1 1 sin x (y = (e−x + e−2x ) ln(ex + 1) + C1 e−x + C2 e−2x ) (y = (C1 + ln | sin x|) sin x + (C2 − x) cos x) 5