《常微分方程》_(方道元_著)_课后习题答案__浙江大学出版社

Bc5B51.Bod1000/H7S_oV:v0+)=0t=T=0BVh = 12C,Ci=V12C = 0,VT = Kh(0) = C = BVZ=cA/H=cA/H/HAi5VSE?SKcAibk)AiM$!AiP(t),5iV8:) = 0,y(0) = 04.20043o81508Hb,T/o88EsYcscssYo88BkB85r!8sYmc,8s,58:(x4, 1 1y(p1aC |p| :y = a(1+2Yy = Cx+C,ii)!t05a 0iK = K()nKHuxB|f(,)f(x,n(x)| 17nKH|n(x)| f(x0,f(x,n(x)|f(,() +f(,()2f(x,n(x)|f(,() +f(,()2f(x,n(x)|+n b/+B/!BfKuB5IBBl0biIKuWHBb!I =a,b)bHqfBf(x)=n(x)Va,bfbf(x)iVb|Bxn,n =1,2,.,Pa (b)y(x)g(x)=)y(x) g(x)+s)g(s)e)b)y(x) g(x)+s)g(s)e)g(x)+g(x)s)e)g(x)(1 +s)e) g(x)(1) g(x)a)ka,PZ= 4()lQb)|k:11rl,1.:(a)!f(x)=x14(),5f0(x)=11P|f0(x)| 0,iBx = ,Px = b)0 0x,y)2R|f(x,y)|, = a,(1);(x0,6= (0,0);L(x0,y =0US;a |5V7 = a,b(|a)2,5NHiuW(2)(0,2(2);aV|av=%GH,B%M?k.:.sZ0N7%f(x,y)sZkZy)3(x0,KviuW1)0ib5T9ff(y)=y)3(x,y)V75BbAy 0,y 1Z+(x0,y =0/ZbwL|qv,5wL9?0V7ZVKviuW1)bfVs!5(E):y3)e(x+y)2,y(KviuWa 0,y = y(x)9,? 1,5,b = 1.(2)(x0,y = 1y =3Wy = 1y =3.Va = 1,b= 1.(3)(x0,y =3Z.(1)V,a = L!a(x)b(x)uLsZa(x)y +b(x), (By = y(x)(Kv)iuWIb:!uW(a1,a1, , I,5ff(x,y)=a(x)y+b(x)u(x,y)|g(y) 0.7H(x)=),)=)i.p(1)TG(H(a),5y(x)0 xal;(2)TG(H(a),5y(x)0 xH(a)5G(y(a) 0G(y(x)0 x aly(x)0 x alb(2).TG( H(a)5G(y(a) 0G()y(a)i5y = y(x)A0 x buWa,b,i0 0,in,i(n,n),P|(n,n)(x0, 1n,O(n,n)w (x;n,n)A (x) 0H,.5PwLn,A,5/()Zc(x)00Bby(x,)=y(0) +sy) +sy)y(x,)=1+y7z(x,) =y(x,)5Tz(x,)=1+sy)zx(x,)=xy)z, z(0) = 1yNz(x,) = sy) 0By(x,) 3.nLsZF5kiiBbiBLsZA(t)X+B(t)uWatbOBHqX(X = X(t)a,b,n,X(t)=(x1(t),xn(t)T,t 2 a,bn_f#A(t),B(t)sYnnLnL_fO1tb_X(t)|X(t)| =|xi(t)|A|A(t)| =j=1|t)|bLsZFS5NsZX(t)=(s)X(s)+B(s) Q/EpsZXn(t)=(s)(s)+B(s) ds n =1,2,yN(t)5Xn(t)9V7B/_f(t)7A =a,bA(t)5|Xn(t)(t)| A(s)|(s)(s)|B,E|Xn(t)(t)| (A(ba)8n!1HXn(t)lX(t)iOX(t)1a,bBOX(t)LsZFb/B!X(t),Y(t)S55X(t)Y(t)=(s)(X(s)Y(s)X(t)B,EV|X(t)Y(t)| t0,t|X(t)Y(t)|A(tn!1VX(t)=Y(t)iW%sZFInsZF5fk(x,yk(k =1,2,n)L!fk =1,2,n)u|x , |yk =VZl)555|x7ff(x,y,(x,y) 2G, |f(x,y,b5ZVswLPZVHb4441.!fa(x)0 :x,0 x 1,y(x); x 1,y= 0;x,0 x 1,y (x)b5n(x) (x)b(2)(1)VM!5(E):x2y2)f(x,y),y(ff(x,y)Oyf(x,y) 0,y 6=0H.5(x0,),5y( 0,y = y(x),),X 0,|y( !1;y(Py(y1(7 = x 0.si,r0(r(x1) Z,x1)r0(x)=f(x,y)f(x, 0(r = y0)34ir0( 0.PAB.x:t)x1+t)x2+t)x3,t)x1+t)x2+t)x3,t)x1+t)x2+t)x3+tQLsZFF(1,1,1)T,1+t,t)T,1,1)kpLsZFY#H) = ) = ) = 0+bQsZFYX(t)=(t)C7(t)=0 +t) X(t)=(22(1 +t)3237+X(t)=(t)1(s)B(s)1(t)=0 110 etetet1+t)=00y0t+1t3t41YX(t)=(t)C +X(t)X(0) = 0HX(t)=0et+t+1(3t)ett3(4t)ett41!X = P(t)etQLZi =1,n)_fP(t)BsQTbp_fFetP(t),et),et)QLZFL1b!A =( (i,j =1,n)t)AE)jP(t)5d(et)et(E +AE)(AE)jP(t)=A(et)yN_fFetP(t),et),et)QLZF/L1!tP(t)+t) +et)et 0t)+t) + t)bLM1!*(t)=11P(t)+t)+Pk(t)P(t)sT*(t)sQj 1)T1V= =0Q37kjQyNV7L1!nnfA1(t)A2(t)ubt)dt=x2(t) 0x1(t)9bTt ! 1t)t_,*x1(t)BHY1byNt ! 1t) ! 0*x1(t) ! 1bt)x1(t)tsvt) 12|t)12x2(t)BHYl0x2(t), (d)Ax1(t)x2(t)8iB,) = 15)=0)=1 )=1)=07x3(t)=x1(t + )5) = )=1) = )=0bBVx1(t+ )=x2(t)yt+ )x2(t+ )=t)x1(t) )x1(t)=x2(t+ ) )x1(t)=x2(t)x1(t)=x1(t+4)H1tpVx2(t)=x2(t+ )!LLZF(/BtFX1(t),X1(t),Xr(t),Xr(t),(t),Xn(t)X(t)VUX(t)_(t),Xn(t)LbkLZFLX = X(t)i,L,X(t)=t)+1(t)+t)+r(t)+(t)+t); (Q,L,(Tf(b(Tf(ZVb!i+i=i =1,r)()ZF/,1,2r+1,1b!+ +|(*)V(1+( +(r+(Xr+ +X1(t),X1(t),Xr(t),Xr(t),(t),Xn(t)FyNL1V7= = = = ,1,2r+1,1,1,2r+1,F*LV,1,2r+1,U(*)V140!dQLsZA(t)X +f(t)f(t) 6=0dQLsZFOn+1L1b!X1(t),(t)dQ75QZ5,LM1bL!X1(t),(t)L1*Y1(t),(t)9L1!/n+1L1!,QFX+57XdQZb+X=0*+)X=0+=0,0 + 6=05X=1+( + *XQZyNf(t)=0f(t) 6=0+=0b=51.p/ZY(1)16 (2)x =0 (3)x =0 (4)x+yz,y +zx,z +xy (5)3x+ 48y28z,4x+ 40y22z,6x+ 57y31z (6)x+4y =0,xy =0 (7)xy +3z,x+z,xy +2z (8)x+5y,5x+3y (9)5x10y20z,x+ 5y + 10z,x+ 4y +9z (10)x = (11)y = (12x7)ex (12)y(4)+2y = y(0) = 1,) = 2,) = 3,) = 0 (13)y =4(14)80)x(1) =13,y(1) = 13 41(15)80)x(1) = 0,y(1) =43b(1)+Z785+ 163=0+0,0,0,2,2,2,2F1,t,t2,e2t,Yc1+t+t(2)+Z4+4=0+(1 +i),(1i)FeeY)+Z443+828 +3=0+1,1,1 Fet,Y)+Z+1,1 +_(1,1,1)T, (1,2,2)T, (1,12)T5Y,1,1)T+1212+22(5)+Z+1,2,3+_(3,2,3)T, (4,1,1)T, (2,2,3)T5Y,2,3)T+,1,1)T+,2,3)T(6)+Z+i,+_(1,12, (1,12,(1,i,12,(1,i,12,5Y,12)T+,12)T+c3(2+c4(2(7)+Z+0,2,3+_(1,3,2)T, (1,1,0)T, (4,3,1)5Y1,3,2)T+1,1,0)T+,3,1)T(8)+Z+3 5i+_(1,i)T, (1,i)T5Y+(9)+Z+5,2 i+_(2,0,1)T, (20 + 10i,155i,142i)T, (2010i,15 + 5i,14 + 2i)T5Y,0,1)T+i)t(20+10i,155i,142i)T+i)t(2010i,15+5i,14+2i)T(10)QZ+Z352+84=0+1,2,2Fet,QZY+ZVA =12yNY21)QZ+Z25 +6=0+2,3FQZY+xZVA =1yNYxex(12)QZ+Z4+22+1=0+i,i,i,iFQZY+ZVA = 18,B= 0yNYSHq5383)QZ+Z22 +2=0+1 FQZY+ZVA =0,B=2yNY4)BZVx =|=ZVy = 5)BZVx =0|=ZVy =t!(t)Z f(t)kff(t)uW0,+1)bk(a)k 6=0HZYVVU(t)=ts) f(s)b)k =0HZYVVUx = c1+t0(ts)f(s)c1,ib(a)k 6=0H0(t)=1ts)f(s)Z f(t)B+7QZ0Yx = 2ZY(t)=2ts) f(s)ds(b)k =0H0(t)=ts)f(s)Zt)=f(t)B+7QZYx = 2t#ZY(t)=2t+ts)f(s)22RZf(t)f(t)(0,+1)!1(t),2(t)ZiO11(t)2(t)ikpaSb7y = 1(t)2(t)5yQZ+Z3+52+ a =01=0,2=5+3=5a2Yy = 2e2t+t11yiO2,30b8a0!a,bf(t)uW(1,+1)fb!(t;)50,x(0) = 0,) = f()Bbkx(t)=t;)5 f(x),x(0) = 0,) = 0bVp0ED/0VAu1,Z(t)u =0L1w1!sZ(t)x =0P(t)t2f7O/T(n+ )22byNbB51.p/H5(1)y = x; y(0) = 2,y(2)=1 (2)y = ex,y(0) = 0,y(1) = 0 46(3)y =0,) +y(1) = 2, 2)y(2) = 0b(1)QZ+Z2+1=0+iFZY+ZA =1yNYxHHqy(x) = 2(12)x(2)QZ+Z23 +2=0+1,2Fex,ZY+ZA = 1yNYHHqy(x)=11e1)EYHHqy(x)=xH58(2)F(s)=3s+4)52s4(3)F(s)=3(s+2)2+92s2(4)F(s)=2s+1)2+16s122.9/fF(s)fIM(1)F(s)=32s4, (2)F(s)=s1(s+1)2(3)F(s)=s+2s+5, (4)F(s)=1(5)F(s)=s3(s4)4, (6)F(s)=52s+10(7)F(s)=s(s2+, (8)F(s)=(1)f(t)=32)f(t)=(12t)e)f(t)=e2)f(t)=12t)(5)f(t)= + 12t+ 24236)f(t)=2e722t)+8e722t)(7)f(t)=8)f(t)=1212e.fMZEp/5(1)y = y(0) = ) = 0;(2)34x = 30x(0) = ) = 0(3)!20x = t, x(0) = ) = 0, !06= 0;(4)y(4)+2y =4y(0) = ) = ) = y(3)(0) = 0.(1)Z|fMi:Lf(t) = Y(s)44Y =Y =2s3(s+4)y(t)=t38e2t+8(2)Z|fMi:Lx(t) = X(s)6 34X =60X =60()(s+34)x(t)=60174(e3t)25e3t)t)+52t)(3)Z|fMi:Lx(t) = X(s) !20X = F0! = !0HX = (20)2x(t)=0t)2!200t)2!0b! 6= !0HX = 2120)!20!2x(t)=0t)!0t)!0(!20!2)b(4)Z|fMi:Lf(t) = Y(s)2Y =Y =4s2()y(t)=2tB/whb/8VTBED0MKEDRh5/80mkb/8MX(t)hMx(t)Z80) (2)(2x+ 1)2(2x+ 1)y =0 (3)x =11+(4)t2(t+ 1)t(2 + 4t+t2)(2 +4t+t2)x = (5)2t+ 1)t+ 1)x =(t2+t1)(6)(1t2)b(1)7x = 53y =0+Z2+4+13 = 0 = 23iy = e2t(y =2)72x+1= t,5y =07t = 5+Z(1)2+2=0 =1,2by = x+ 1) +x+ 1)2.(3)QZ+Z2+1=0QZYx(t)= MsTx(t)=ts)1+4)n5pZ+Tx = cZa = 1, c =0b+x = btbNx = tMQZbEpQZY!x = tu(t)Zt+1)t+2)07v = 5v = C(t+1)u = KZYx = 1t+5)n5pZ+Tx =(at+b)Za =1b =0N+50bpQZL1iQZ2t+ 1)t+ 1)x = t(t+ 1)(x)=0,+E!x = C(t)ZbKZx =(2t2)et+6)7y =5Z(1t2)y =0)Ep!y =Z5an=n3bYy = 12k 2(2k3)!(2k)!bNZYx = c0+c2(t16k 2(2k3)!(2k)!2k +1)p/sZx = Z7L1)iMwT(1)y =0,0; (2)y =0,1;(3)(1x)y =0,0; (4)0,0;(5)0,0; (6)(y =0,.(1)7y =ZPn(n + 2)(n + 1) an)wnbTsY|(a0,= (1,0)(a0,= (0,1)VL1k0k)!,k0(2k + 1)!.(2)7y =an(x1)n|Zx1)y0y0y =0ZPn(n+