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Chapter 7 Higher Order Linear Differential Equations In this section we generalize some of the techniques for solving second order linear equations discussed in Chapter 5 so that they apply to linear equations of higher order. Recall from Chapter 4 that an nth order linear differential equation has the form yo) + Pi(a)y)) + Po(w)y9) + ++ + Pally = On) (71) The general solution is y= Cin + Crys + +--+ Cryn + Yps where 41, yo; --+;Yn are linearly independent solutions to the auxiliary homoge- neous equation and yp is a solution of Equation (7.1). So given 1,42, -.-; Yn; how do we find yp? We again consider the two meth- ods we have looked at in §5.2 and §5.3 (p. 60 and p. 66, respectively). 7.1 Undetermined Coefficients If Pix), Pa(or) mined coefficients. .P,() are constants and R() is “nice”, we can use undeter- Example 7.1. Solve y\) —y = 2c +3e”. * 79 80CHAPTER 7. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS Solution. We have pA) = At = 1 =(A- 1) (AF 1) (7 +1). Therefore, ya = sin(e). ywre, yo=e", y3 = cos( Let yp = Ae*” + Bre*”. Then by the short cut method (§5.2.1, p. 64), we have ava! 22 pQ) 1-1 15 3 3 3 B= =_"_=*. pl) 4-3 4 Therefore, the general solution is x 2 3 y = Cre” + Coe* +C3 cos(x) + Cy sin(x) + ye + Gee . > 7.2 Variation of Parameters Suppose that y1,y2,...,Yn are solutions of the auxiliary homogeneous equation. We look for solutions of Equation (7.1) of the form y= vy + v2Y2 T+ + UnYny where v1, 0,...,Un are functions of x, One condition on v1, v2,..., Un is given by substituting into Equation (7.1). We choose n — 1 conditions to be oly + by +++ + Uh Yn = 0, (1) uly) + Upya + + URN = 0, (2) vty?) + oy?) + spake = 0. (n-1) 82CHAPTER 7. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS Conditions (1),...,(n — 1) can be written as vt 0 vi, 0 ul l=]. of R(2) where wa Yn uh Uy M= : y” 1) (n—1) i Therefore, || = W 40. 7.3 Substitutions: Euler’s Equation Consider the equation aty™ faye tyl + tan iy! + day = R(x), (7.2) where a1, a,...,@y are constants. Consider x > 0. Let « =e" so that u = In(x) (for a <0, we must use u = In(—a)). Then dy dydu _ dy1 dz dudx duz’ Py yl dyl dx? du2 a dua? -(%5- t) 5. 2 a dy 373 @y g ot- (3 ye F2) gS 2 _(@y _ uy , ydy -(4 wr +2) 5, ale ay [dy aly dy\ 1 =-(@%.1¢,5 44... dan ( Vat t 7.3, SUBSTITUTIONS: EULER’S EQUATION 83 for some constants C), C2,...,Cn. Therefore, Equation (7.2) becomes ay gt . a ee +n + bay = R(e") for some constants b1,b2,...,Dn- Example 7.2. Solve xy" + ay! —y =a for x > 0. * Solution. Let «= e" so that u = In(a). Then dy _dydu_ dy dw dude dua” &y_ @yl dy (2 #) 2 de dew dux Therefore, dy dy dy au (So-P) + Bava, Pu du? -y=e Considering the homogencous case, we have \?— 1=0 + \ = +41. Therefore, —". Using undetermined coefficients, let yp = Ae". Then = 9Ae3", Substitution gives yi = e" and yo y! = 3Ae™ and y" 9 Ae — Ac =e, ecminle, BAER We can immediately see that A= 1/8. Therefore, the general solution is -u, los y= Cre + Coe™ + oe = Cye™®) 4 Cue PO) 4 zon) Cy, 2 =Cw+ = ae where C; and Cy are arbitrary constants. > We can also consider a direct method. Solving dy dy Ie +p +u=0 
