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Chapter 5 Second Order Linear Equations A second order linear equation is an equation of the form y" + Pla)y! + Q(a)y = R(x). (5.1) To find the general solution of Equation (5.1), we need * Two solutions: y; and yy of the corresponding homogeneous equation y' +P(a)y! + Q(x) =0. © One solution: yp of Equation (5.1). Then the general solution is y = c1y1 +c2yo+ yp, where ¢, and cg are arbitrary constants. Except in the case where P(a) and Q(«) are constants (considered earlier on p.52), there is no general method of finding yi, but L. given y1, there exists a method of finding yo. 2. given y, and yo, there exists a method of finding yp. We will discuss (1) first. 5.1 Reduction of Order As mentioned earlier, we can find a second solution from the first. We first develop the tools we need. 58 CHAPTER 5. SECOND ORDER LINEAR EQUATIONS Lemma 5.1 Let f(a) be a twice differentiable function on a closed bounded interval J. Ii{«w € J: f(x) =0} is infinite, then there exists an a € J such that f(x) =0.and f'(ao) =0. Proof Let S = {rE J: f(x) =0}. If [S| is infinite, then by the Bolzano- Weierstrass Theorem, 5 has an accumulation point a, ie., there exists an ao € S such that every interval about ao contains another point of $. Therefore, by Rolle’s Theorem, every interval about 2p contains a point where f is 0. Therefore, by continuity, f’(#o) = 0." a Corollary 5.2 Let yi(a) be a solution of y" + P(x)y’ + Q(x)y = 0 on a closed bounded interval J. If y; is not the zero solution, then y;(«) = 0 for at most finitely many x in J. Proof. Let V be the set of solutions of y" + P(a)y! + Q(a)y = 0 on J. Then V is a two-dimensional vector space. So if y: #0, then there exists a yo € V such that y; and yp are linearly independent. Therefore, W(y1,y2) # 0, ie., W(y1,42) is never zero (the Wronskian is either always 0 or never 0). But if yi(x) = 0 for infinitely many a in J, by Lemma 5.1, there exists an xo € J such that yi(o) =0 and yf(ao) = 0. Therefore, 0 y2(x0) 9 y5(ao0) W (41, y2)(«0) = yi(xo) y2(xo) y\(wo) y3(ao) |-° This is a contradiction. Therefore, y;(«) = 0 for at most finitely many # in J. Let y; a solution to y"” + P(x)y’ + Q(a)y =0 on a closed bounded interval J. We want to find a linearly independent second solution yp. We break J into a finite collection of subintervals so that y,(«) 4 0 on the interior of each subinterval. We will describe how to find the second solution y2 “Note that f’ is necessarily—but certainly not sufficiently—continuous if f" is differen- tiable. 60 CHAPTER 5. SECOND ORDER LINEAR EQUATIONS Example 5.3. Observe that y = 2 is a solution of 1 @H2,, 242 a y+ a-y=0 and solve 9 9 ee . Solution. Using the fact that y, = x, we have ete pe J (- 84 ae ef (IE) de ert2in((2|) gyn cane [eee s ea er e2 (lel) e* x 2 == = demo [ 2 tema [ede = ae Therefore, x ae” a Wel) 284 met | = pare? — ae = a7 ef +2 From this, we have wR wetae® = Be = AO = -e Se =e", R_ axe™ vy = ee ni wae. Ww arer Therefore, 2 Yp = iy + vay2 = —axe" + ae", and the general solution is y = Cie + Cae” — xe” +a%e" = Cir t+ Chae +02", a £0, where C1, C, and Cf, are arbitrary constants. © 5.2 Undetermined Coefficients Suppose that y; and yp are solutions of y+ P(«)y! + Q(«)y = 0. The general method for finding yp is called variation of parameters. We will first consider another method, the method of undetermined coefficients, which does not always work, but it is simpler than variation of parameters in cases where it does work. 5.2. UNDETERMINED COEFFICIENTS 61 The method of undetermined coefficients has two requirements, that 1. wo have constant cooflicients, i.e., P(e) = p and Q(z) = q for some con- stants p and g, 2. R(x) is “nice”, Example 5.4. Solve y — y=". * Solution. We have \2 —1 = 0. So 41 =e and yo =e7*. Since R(x) = 2", we look for a solution of the form y = Ac**, Then y’ = 2Ae**, y" = 4Ae**. Then according to the differential equation, we have 4A” — Ae? = SS BAe? . 1 pT Hs AK, 3 Therefore, the general solution is 2 -n24 12 y=ce” + ee Wege where c; and cy are arbitrary constants. % Undetermined coefficients works when P(«) and Q(«) are constants and R(x) isa sum of the terms of the forms shown in Table 5.1 Table 5.1; The various forms of yp ought to be given a particular R(e). Re) Form of ys Ce" Aga? + Ayah} 4... + Ane t+ An Cev™ Ae™™ Coe saa Ae? cos(kat) + Be" sin(ke) The success of undetermined coefficients depends on the derivative of R(«) having some basic form as R(x). There is an easy way to deal with a complica- tion: whenever a term in the trial yp is already part of the general solution to the homogeneous equation, multiply it by x. For example, if we have y” — y =e", as before, we have y1 =e" and yp = e2". Thus, we try yp = Awe?” and find A. 5.2. UNDETERMINED COEFFICIENTS Therefore, yp = —2cos(x)/2, and the general solution is : 7 cos y= exsin() + ¢2.005(2) =, where c, and c2 are arbitrary constants. Example 5.7. Solve y"” — dy = a. 63 © * Solution. We have \? — 4 = 0, so. \ = +2, and it follows that y, = e2* and yo =e". Let y = Ae®+ Ba+C. Then y’ = 2Ax+B and y” =2A. Therefore, y" — 4y = 2A — 4A? — 4B — 40 = —4Ax? + 0a + (2A— 4B -4C). Comparing coefficients with a”, we immediately see that 1 AAS 1 A=—7, 4B =0= B=0, A_-1/ft 1 2A-AC = 09 C= Fa =. Therefore, yp = —a?/4 — 1/8, and the general solution is Qa 20 1 = cie™™ + cge 7* — — — =, y 1 2 7 3 where ¢; and cy are arbitrary constants. Example 5.8. Solve y"” — y'! =a. ) * Solution. We have A? — A = 0, so \ € {1,0}, and it follows that y; = e” and yo = 1. Let y = Ax? + Bx. Then y' = 2Aa+ B and y" = 2A. Therefore, y! —y =2A—2Aw - B, Comparing coefficients with x, we immediately see that 1 tA=1l=> A=—5, 2A-B=0=>B=-1. Therefore, yp = —a?/2 — a, and the general solution is Fo y=ce+e—->— a, 64 CHAPTER 5. SECOND ORDER LINEAR EQUATIONS where cy and ¢» are arbitrary constants. © 5.2.1 Shortcuts for Undetermined Coefficients Suppose we have y" + ay’ + by = R(x). Let p(A) =? + a\+ b. We now consider two cases for R(.). Case I: R(x) = ce. It follows that yp = Ae, so y! = ade and y”" =aAe™, Then yl + ay! + by = ce (a? + aa +b) Ae*® = pla) Ae*™ Therefore, A = c/p(a), unless p(a) =e. What if p(a) = 0? Then e°* is a solution to the homogeneous equation. Let y= Awe, Then y! = Ac*® + Aame®®, yl! = Ane?” + Ave? + Aatne™ =2Ane® + Aa2ne Thus, y" +y' +y=Al[(a2e + aax +b) +2a+a] = A(p(a)we™ + p'(a)e**) = Ap'(a)e%*. Therefore, A =c/p'(a), unless p!(a) =0 too. If p(a) = 0 and p'(a) = 0, then e* and ae*” both solve the homogeneous 208 equation. Letting y = Axe", we have A = c/p"(a). Case Il: R(«) = ce" cos(br) or ce" sin(bx). We use complex numbers. Set a=r+ib. Then ce” = ce" cos(be) + ice™ sin(be). Let x = y +iw. Consider the equation 2! faz! tbe =ce™. (5.3) 66 CHAPTER 5. SECOND ORDER LINEAR EQUATIONS 5.3 Variation of Parameters Consider the equation y" + P(@)y' + Q(a)y = R(x), () where P(x) and Q(a) are not necessarily constant. Let yy and yp be linearly independent solutions of the auxiliary equation oy! +P(a)y! + Q(x)y = 0. (H) So for any constants Cy and Cy, Cry, + Coyp satisfies Equation (H). The idea of variation of parameters is to look for functions v;(z) and v(x) such that v1y) + v2yp satisfies Equation (1). Let yy and y. If we want y to be a solution to Equation (1), then substitution into Equation (1) will give one condition on vy and v2. Since there is only one condition to be satisfied and two functions to be found, there is lots of choice, i.c., there are many pairs vy) and vy which will do. So we will arbitrarily decide only to look for v1 and v2 which also satisfy v{y +vsy2 = 0. We will see that this satisfaction condition simplifies things. So consider y= vy + v2y2, y= vy + vays + vty + vay2 = ery} + v2V>, y= vy + vayd + oly + vbyh. To solve Equation (I) means that vin +eays + vtyh + uh + Pleat) _ R(e) +P(x)voy + Q(z)viy + Q(w)v2ye v1 (yt + P(e) + Q(e)n) +2 (uy + P(x +Q(x)y2) | = R@)- tuly, + vbu5 Therefore, vin + May = R(x), fg tosh vyyy + Vay = 0, 5.3. VARIATION OF PARAMETERS 67 and it follows that vf = wk ean ; ae WwW a Ww where #2 | 20, Yo ie., it is never zero regardless of a, since y; and yp are linearly independent. Example 5.10. Solve y" — 5y' + 6y =a7e**, * Solution. The auxiliary equation is yf! —5y! + Gy = 0 \? -5A4+ 650A =3,2. Therefore, the solutions are y; = ¢3” and yy = 2". Now, 3x 2x , a e seas 7 waft Bll] CO oT | Hach — gefe = pb vi | | Be 2 Therefore, . E28 72 3r 3 ee We can use any function v; with the right v{, so choose the constant of integra- tion to be 0. For v2, we have so w= fee die = —a7e" + 2f axe” da by parts by parts = ae? + Qre™ — 2 [ erde = ae" + Qe” — 2e* = — (uv? — 2x + 2) e. Finally, 2 r = Bon a (# = w+ 2) ere?" = (F-2+22-2), 5.3. VARIATION OF PARAMETERS 69 Finally, Yp = viy + v2y2 cos(e*) + €~2* (€” cos(e*) — sin(e*)) ® cos(e”) + €-” cos(e”) — € ?* sin(e”) =e?" sin(e”) , and the general solution is Qa y= Cie* + Coe” — €** sin(e*) , where C; and C) are arbitrary constants. co) 
