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Solving linear, const.-coeff. ODEs This is a summary of the method for finding the general solution of linear, first and second order, constant-coefficient ODEs. Details of the theory can be found in most introductory books on ordinary differential equations such as Boyce & DiPrima. The ODE to be solved for y = y(x) is of the form a y 00 + b y 0 + c y = F (x) , (1) where a , b and c are real constants (thus the classification as a constant-coefficient equation), and F (x) is a given (forcing) function. Note: If F (x) ≡ 0 the equation is homogeneous, otherwise it is non-homogeneous. 1 First order equations If a = 0 and b 6= 0 , then the equation is a first order ODE b y 0 + c y = F (x) , (2) and its general solution is 1 x −c(x−ξ)/b −c(x−x )/b 0 y(x) = y(x0) e + e F (ξ) dξ , b x0 where x0 can be chosen arbitrarily. Z This is obtained multiplying (2) by the integrating factor 1 cx/b e . b It can also be obtained by the method described next for µ(x) = 2nd order equations. Homogeneous 2nd order equations If a 6= 0 the ODE is 2nd order, and the general solution of (1) can be decomposed into the sum of the general solution of the corresponding homogeneous equation, a y 00 + b y 0 + c y = 0 , (3) and one particular solution of the non-homogeneous equation (1). Thus we have y(x) = yh(x) + yp(x) , where yh(x) is the general solution of (3), and yp(x) is a solution of (1). In this section a method to find yh is presented, and in subsequent sections methods to find yp will be discussed — they can also be found in most introductory books. Variation of parameters, and the method of undetermined coefficients will be reviewed in this document. Assume that the solution of (3) has the form y(x) = erx and plug into (3) to conclude that r must be a root of the characteristic equation a r2 + b r + c = 0 . (4) The form of yh depends on the roots of (4). The 3 possible cases will be discussed next. Real, distinct roots Suppose b2 − 4ac > 0 . In this case there are two distinct real roots of the characteristic equation, r 2 b − 4ac b r1,2 = − ± , 2a 2a and the general solution of the homogeneous equation (3) is yh(x) = A1 er1x + A2 er2x , where A1 and A2 are arbitrary constants which are usually determined from initial and/or boundary conditions. An alternative form of the general solution of (3) which is sometimes preferred is yh(x) = eλx {B1 cosh(µx) + B2 sinh(µx)} , (5) where B1 and B2 are arbitrary constants, and b λ=− , 2a µ= r 2 b − 4ac 2a , r1,2 = λ ± µ . (6) As an example, consider the following ODE y 00 − y 0 − 2y = 0 which has characteristic equation r2 −r−2 = 0 , with roots r1 = 2 , r2 = −1 . The general solution of the ODE can be written as y(x) = A1 e2x + A2 e−x , or as (here λ = 1/2 , µ = 3/2 and r1,2 = λ ± µ ) y(x) = ex/2 {B1 cosh(3x/2) + B2 sinh(3x/2)} . Real, equal roots If b2 − 4ac = 0 , then the roots of the characteristic equation are b , 2a and the general solution of (3) is r1 = r2 = λ = − yh(x) = eλx {A1 x + A2} . As an example consider the ODE y 00 + 4y 0 + 4y = 0 . The characteristic equation r2 + 4r + 4 = 0 has only one root: r1 = r2 = λ = −2 , and the general solution of the ODE is y(x) = e−2x {A1 x + A2} . Complex conjugate roots If b2 − 4ac < 0 , then the roots of the characteristic equation are complex conjugate numbers r1,2 = λ ± iµ , where λ and µ are given in (6). In this case, the general solution of the homogeneous equation (3) is (note the similarity with (5)) yh(x) = eλx {A1 cos(µx) + A2 sin(µx)} . As an example consider the ODE y 00 + 4y 0 + 13y = 0 . The roots of the char. equa. are r1,2 = −2 ± 3i (here λ = −2 , µ = 3), and the general solution is y(x) = e−2x {A1 cos(3x) + A2 sin(3x)} . Constant-coefficient case: Method of undetermined coefficients Let L[y] := ay 00 + by 0 + cy . The goal is to find a particular solution yp(x) to the non-homogeneous equation (1) L[y] = F (x) , ay 00 + by 0 + cy = F (x) . i.e. In order to do this, note that if the forcing function is a sum F (x) = m X Fj (x) , (7) j=1 then we can decompose yp(x) as a sum yp(x) = m X ypj (x) , (8) j=1 where L[ypj ] = Fj (x) , j = 1, . . . , m . (9) Indeed, from (8),(9),(7) and the linearity of the ODE we obtain L[yp] = L m X j=1 ypj = m X j=1 L[ypj ] = m X j=1 Fj (x) = F (x) . The method of undetermined coefficients works for the case of constant-coefficient, linear ODEs, when the forcing function is of a special form as described below. Assume that or F (x) = Pn(x) eα x cos(β x) , (10) F (x) = Pn(x) eα x sin(β x) , (11) where Pn(x) is a polynomial of degree n , and α , β are real numbers. Assume that α+iβ is a root of multiplicity s of the characteristic equation ar 2 + br + c = 0 , i.e. • s = 0 means that α + iβ is not a root of the characteristic equation, • if s = 1 then α + iβ is a simple root of the characteristic equation, • if s = 2 then α + iβ is a double root of the characteristic equation. In the cases (10),(11), the form of the particular solution is yp(x) = xs eα x [Qn(x) cos(β x) + Rn(x) sin(β x)] , where Qn(x) and Rn(x) are polynomials of degree n (same degree as Pn(x)) with coefficients to be determined from (1), i.e. Qn(x) = An xn + An−1 xn−1 + · · · + A1 x + A0 , Rn(x) = Bn xn + Bn−1 xn−1 + · · · + B1 x + B0 , where An, . . . , A0 and Bn, . . . , B0 have to be determined from (1). Note some special cases: (a) If α = β = 0, then s is the multiplicity of 0 (zero) as a root of the characteristic equation; (b) if β = 0, then s is the multiplicity of α as a root of the characteristic equation; (c) if α = 0, then s is the multiplicity of iβ as a root of the characteristic equation. The following examples will illustrate the method. Example 1 L[y] := y 00 + 5y 0 + 6y = 4ex cos(2x) (12) The char. equa. has roots r1 = −2 , r2 = −3. Therefore, 1+2i is not a root of the char. equa., and s = 0. So, we should try (here α = 1, β = 2, n = 0 in (10)) yp(x) = ex [A0 cos(2x) + B0 sin(2x)] . Then, yp0 (x) = ex [(A0 + 2B0) cos(2x) + (B0 − 2A0) sin(2x)] yp00(x) = ex [(4B0 − 3A0) cos(2x) −(4A0 + 3B0) sin(2x)] , so that L[yp] = ex [(8A0 + 14B0) cos(2x) − (14A0 − 8B0) sin(2x)] . Thus, yp satisfies (12) provided 8A0 + 14B0 = 4 14A0 − 8B0 = 0 i.e. A0 = 8/65 and B0 = 14/65 , and ex yp(x) = [8 cos(2x) + 14 sin(2x)] . 65 (13) Example 2 L[y] := y 00 + 5y 0 + 6y = −5x sin(3x) (14) The char. equa. has roots r1 = −2 , r2 = −3, and 3i is not a root of the char. equa., so s = 0 and we should try (here α = 0, β = 3, n = 1) yp(x) = (A1x+A0) cos(3x)+(B1x+B0) sin(3x) . Then, yp0 (x) = (A1 + 3B1x + 3B0) cos(3x) + (B1 − 3A1x − 3A0) sin(3x) yp00(x) = (6B1 − 9A1x − 9A0) cos(3x) − (6A1 + 9B1x + 9B0) sin(3x) , so that L[yp] = [(15B1 − 3A1)x + (5A1 + 6B1 − 3A0 +15B0)] cos(3x) − [(15A1 + 3B1)x + (6A1 − 5B1 +15A0 + 3B0)] sin(3x) . Thus, yp satisfies (14) provided 15B1 − 3A1 5A1 + 6B1 − 3A0 + 15B0 15A1 + 3B1 6A1 − 5B1 + 15A0 + 3B0 =0 =0 =5 =0 i.e. A0 = −235/3042 , B0 = −25/169 A1 = 25/78 , B1 = 5/78 , and 235 25 x− cos(3x) yp(x) = 78 3042 5 25 x− sin(3x) . (15) + 78 169 Example 3 L[y] := y 00 + 5y 0 + 6y = 2x2e−3x (16) The char. equa. has roots r1 = −2 , r2 = −3, and −3 is a simple root of the char. equa., so s = 1 and we should try (α = −3, β = 0, n = 2) yp(x) = x (A2 x2 + A1 x + A0) e−3x . Then, h 0 yp(x) = (−3A2 x3 + (3A2 − 3A1)x2 + (2A1 − 3A0)x i + A0 e−3x h 00 yp (x) = 9A2 x3 − (18A2 − 9A1)x2 i + (6A2 − 12A1 + 9A0)x + (2A1 − 6A0) e−3x so that h L[yp] = − 3A2 x2 + (6A2 − 2A1)x + (2A1 − A0) e−3x . i Thus, yp satisfies (16) provided −3A2 = 2 6A2 − 2A1 = 0 2A1 − A0 = 0 i.e. A0 = −4 , A1 = −2 A2 = −2/3 , and 2 2 yp(x) = −x x + 2 x + 4 e−3x . 3 Example 4 (17) L[y] := y 00 + 5y 0 + 6y = 2x2e−3x − 5x sin(3x) + 4ex cos(2x) Since the forcing function here is a sum of the forcing functions in (12),(14) and (16), we conclude from (8),(13),(15) and (17) that 2 2 x + 2 x + 4 e−3x yp(x) = −x 3 25 235 + x− cos(3x) 78 3042 5 25 + x− sin(3x) 78 169 ex + [8 cos(2x) + 14 sin(2x)] . 65 General case: Method of variation of parameters This method relies on the knowledge of a fundamental set of solutions of the homogeneous linear equation y 00 + p(t) y 0 + q(t) y = 0 , say { y1(t), y2(t) }, to obtain a particular solution of the non-homogeneous equation (in normal form) y 00 + p(t) y 0 + q(t) y = F (t) . (18) The idea is to postulate a form of the particular solution as yp(t) = u(t)y1(t) + v(t)y2(t) , (19) and to find the (variable) parameters u and v so that yp(t) satisfies (18). The ODE gives one condition on the two parameters. We are free to impose an additional condition for these two unknowns. The following system of equations is obtained. y1 u0 + y2 v 0 = 0 y10 u0 + y20 v 0 = F (t) (20) (21) Since the coefficient matrix of this algebraic system has determinant equal to the Wronskian of the fundamental set (which is nonzero) y 1 y2 0 − y0 y , = y y W (t) := W [y1, y2] = 0 1 0 2 1 2 y1 y2 there is a unique solution for u0 and v 0 , which can be expressed (Cramer’s rule) as u0(t) = v 0(t) = 0 y2 0 F (t) y2 W (t) y 0 1 0 y1 F (t) W (t) −y2(t)F (t) = , W (t) y1(t)F (t) = . W (t) Hence, u(t) = − v(t) = Z Z y2(t)F (t) dt , W (t) y1(t)F (t) dt . W (t) (22) (23) Example 5 t2y 00 − 3ty 0 + 4y = t2 ln t , t > 0. (24) It is easy to verify that y1 = t2 and y2 = t2 ln t form a fundamental set of solutions of the corresponding homogeneous ODE on the interval (0, ∞). Also, the normal form of (24) is 3 0 4 00 y − y + 2 y = ln t , t t t > 0. Therefore, the method of variation of parameters gives — see (19) yp = u y 1 + v y 2 , where u0 and v 0 satisfy the system of algebraic equations (20),(21) t2 u0 + t2(ln t) v 0 = 0 , 2t u0 + t(1 + 2 ln t) v 0 = ln t , which has the solution (22),(23) (you can solve the system above any way you want) (ln t)2 0 u =− , t Upon integration, we obtain (ln t)3 u(t) = − , 3 v0 = ln t . t (ln t)2 v(t) = , 2 so that t2(ln t)3 t2(ln t)3 + yp(t) = − 3 2 t2(ln t)3 or yp(t) = . 6 Thus, the general solution of (24) is t2 (ln t)3 2 2 y = c1 t + c2 t ln t + , 6 or y = t2 c1 + c2 ln t + (ln t)3 6 ! .