二階常系數(shù)非齊次線性微分方程的特殊解法

1、2003年5月第9卷第2期安慶師范學院學報(自然科學版Journal of Anqing Teachers College(Natural ScienceMay.2003Vol.9NO.2二階常系數(shù)非齊次線性微分方程的特殊解法舒阿秀1,何家慧2(1安慶師范學院數(shù)學計算機系,安徽安慶246011;2安徽省體校,安徽合肥230000摘要:本文主要介紹幾種不同類型的二階常系數(shù)非齊次線性微分方程的三種相對簡捷的解法。關(guān)鍵詞:微分方程;特解;通解中圖分類號:O175.1文獻標識碼:A文章編號:1007-4260(200302-0103-03對于二階常系數(shù)非齊次線性微分方程:y+p y+qy=f(x的解法

2、,一般教科書上主要歸結(jié)為其特解的求法,并針對f(x的不同類型分為常數(shù)變易法與待定系數(shù)法兩種。但往往求解過程較為繁瑣,計算量較大。本文主要針對上式中f(x的不同類型介紹幾種相對靈活簡便的求解方法。一、升降法1.先考慮f(x=P m(x,(P m(x為m次多項式的情形。設(shè)f(x=a m x m+a m-1x m-1+a1x+a0,則y+p y+qy=a m x m+a m-1x m-1+a1x+a0(1. 1對上式兩邊連續(xù)求導m次,直至方程右邊為一個常數(shù),可得:y+p y+qy=ma m x m-1+(m-1a(m-1x m-2+a1y(m+1+py(m+qy(m-1=a m m(m-12x+a

3、m-1(m-1!y(m+2+py(m+1+qy(m=a m m!在上面一系列式子的最后一個中,令y(m=a mq m!(此時y(m+2=y(m+1=0,逐次由下向上代入直至(1.1式即可推得(1.1的一個特解。例1求y-6y+9y=x3的特解。解對原方程連續(xù)求3次導數(shù)得:y-6y+9y=3x2;y(4-6y+9y=6x;y(5-6y(4+9y=6令y=6/9,(此時y(5=y(4=0逐個代入可得:y=23x +49;y=13x2+49x+29;y=19x3+29x2+2 9x+881。2.設(shè)f(x=P m(xe x(P m(x為m次多項式,則y+py+qy=P m(xe x(1.2,此時可令y

4、(x=z(xe x代入(1.2并整理得:z+p1z+q1z=Q m(x,(p1,q1為常數(shù),Q m(x為m次多項式,從而歸為1的類型。例2求y-2y=4x2e2x的特解。解令y(x=z(xe2x代入并整理得:z+2z=4x2,利用升降法可得:z+2z=8x;z(4+2z=8再令z=4,(此時z(4=0得:z=4x-2,z=2x2-2x+1,從而z=23x3-x2+x,所以所求特解為:y作者簡介:舒阿秀(1977-,女,安徽旌德人,安慶師范學院數(shù)學計算機系教師,從事代數(shù)教學。收稿日期:2002-12-24=(23x 3-x 2+x e 2x 3.設(shè)f (x =P (x cos x +Q (x s

5、in x e ax ,(P (x ,Q (x 為多項式,則y +py +qy =P (x co s x +Q (x sin x e x (1.3利用疊加原理,(1.3的特解y (x 等于方程:y +py +qy =P (x e ax cos x 的特解y 1(x 與方程:y +p y +qy =Q (x e ax sin x 的特解y 2(x 之和,而y 1(x 又等于方程:z +p z +qz =P (x e (a +i x 的特解z (x 的實部,即y 1(x =R e (z (x ;y 2(x 又等于方程:z +p z +qz =Q (x e (a +i x 的特解z (x 的虛部,即y

6、 2(x =Im (z (x 。同時z (x 的求解可歸為2的類型,所以(1.3的特解也可用升降法求得。例3求y -2y +2y =x e x sin x 的特解。解先求方程z -2z +2z =x e (1+i x 的特解。令z (x =u (x e (1+i x ,代入得:u +2iu =x ,用升降法求得其特解為:u =-i 4x 2+14x 于是z (x =(-i 4x 2+14x e (1+i x =(14x cos x +14x 2sin x e x +i (14x sin x -14x 2cos x e x ,從而所求特解為:y (x =I m (z (x =(14x sin x

7、 -14x 2cos x e x 。二、公式法設(shè)(1對應的齊次線性方程的特征根為 1, 2,由韋達定理:p =-( 1+ 2,q = 1 2則(1化為:y -( 1+ 2y + 1 2y =f (x ,即(y - 1y - 2(y - 1y =f (x 。若令y - 1y =z ,則有z - 2z =f (x ,這是一階線性方程,易解得:z =e 2x f (x e - 2x d x ,從而z =e 1x e ( 2- 1x f (x e - 2x d x d x ,(2.1對(2.1進一步利用分部積分運算并令積分常數(shù)為零可得到以下關(guān)于(1的特解的公式:(i若 1 2,則y =1 2- 1e

8、2x e - 2x f (x d x -e 1x e - 1x f (x d x (2.2特別地,若 1,2=a ±i ,則y =e ax sin x e -ax f (x cos x d x -co s x e -a x f (x sin x d x (2.3(ii若 1= 2,則y =e 1x x e - 1x f (x d x -x e - 1x f (x d x (2.4這三個公式的適用范圍較廣而且f (x =e a c cos x ,e ac sin x 時或 1、2為共軛復根時求解過程要比待定系數(shù)法簡捷。例4求y -2y +2y =x e x sin x 的特解。解因y

9、-2y +2y =0的特征根為 1,2=1±i ,所以由(2.3原方程有特解:y =e x sin x e -x x e x sin x cos x d x -cos x e -x x e x sin x sin x d x =e x (sin x x sin x co s x d x -cos x x sin 2x d x =(14x sin x -14x 2cos x +18co s x e x (取積分常數(shù)為零注:若取積分常數(shù)c 1=0,c 2=-1/4,即有y (x =(14x sin x -14x 2cos x e x ,與例3的結(jié)果相同。例5求y -6y +9y =2x

10、e 3x 1+x 2的特解。解因y -6y +9y =0的特征根為1= 2=3,故由(2.4原方程有特解:(取積分常數(shù)為零y =e 3x x e -3x 2x e 3x 1+x 2d x -x e -3x 2x e 3x 1+x 2d x =e 3x x ln(1+x -2x +2arctan x 。三、拼湊法104安慶師范學院學報(自然科學版2003年為簡便起見,僅以f (x =P 2(x e x ,(P 2(x 為二次多項式為例說明此法的可行性。設(shè)f (x =e x (ax 2+bx +c ,則y +py +qy =e x (ax 2+bx +c (3.1若 不是對應齊次線性方程的特征根,

11、即 2+p +q 0時,(3.1可拼湊為:y -e x (a x 2+b x +c +p y -e x (a x 2+b x +c +q y -e x (a x 2+b x +c =0(3.2其中:a =a 2+p +q ;b =b 2+p +q -(4 +2p a ( 2+p +q 2;c =c 2+p +q -2a +(2 +p b ( 2+p +q 2+2a (2 +p 2( +p +q 令y -e x (a x 2+b x +c =z ,則(3.1可化為齊次線性方程:z +p z +qz =0進行求解。例6求y +2y +2y =x 2+2x -1的通解。解由(3.2得:y -(12x

12、 2-1+2y -(12x 2-1+2y -(12x 2-1=0令y -(12x 2-1=z 得:z +2z +2z =0,其通解為:z =(c 1cos x +c 2sin x e -x所以所求通解為:y =(c 1cos x +c 2sin x e -x +12x 2-1。通過例子可以看出,利用此法可避免繁瑣的求特解過程,直接將(1拼湊為齊次線性方程求得通解,但它只適用于 不是對應齊次線性方程的特征根的情形,適用面不廣。同時若 =0或q =0則用此法更為簡便。同樣地,若f (x =P (x cos x +Q (x sin x e ax ,(P (x ,Q (x 為多項式,對于 =a +i

13、不是對應齊次線性方程的特征根的情形也可用此法直接求通解,但相對復雜一些,此處不作介紹。參考文獻1E.R.Love.Particular Solutions of Cons tant coefficien t Linear Differen tial Equations.Th e Ins titu te of M athem atics and its Applica-tion s,25,(6:165166.2胡偉鵬.一類二階常系數(shù)非齊次線性微分方程的解法J.高等數(shù)學研究,2001.4(2:4546.3周尚仁,權(quán)宏順.常微分方程習題集M .北京:人民教育出版社,1980.Particular S

14、olutions of Nonhomogeneous Linear DifferentialEquations with C onstant Coefficient of the Second OrderSHU A -xiu 1;HE Jia-hui 2(1T he M ath .and Compu ter Dept .of Anqin g T each ers 'college ,Anqing 246011,Ch ina2Anh ui Ph ysical Education School ,H efei 230000,Ch ina Abstract :T his paper pr imar ily int ro duces thr ee simple solutio ns of sever al kinds of nonhomo -ge