常微分方程教程课后习题答案第16章

1、11? ? VgSK 1-11.)?(1) y = C1e2x + C2e2x y0 = 2C1e2x 2C2e2xy00 = 4C1e2x + 4C2e2x y00 4y = 0 e2xe2xDy,y0 q= 4 = 06DC1,C2 2e2x2e2x d?m?A? ?)"(2) y =sin x x y0 =x cos xsin xx2xy0 + y = cos x y = sin x ?xy0 + y = cos x ?A)"xR(3) y = x( xe dx + C)1 xR01 xxy =xe dx + C + e xy0 y = xexq= x 6= 0, (

2、x 6= 0)dydcR y = x( xe dx + C)?01 xxy y = xe ?)", < x < c1x12 (x c )41(4) y =, c1 x c2()012(x c ), c2 < x < 241 (x c ) , < x < c121 y0 =0, c1 x c21(x c ), c2 < x < 22y0 = p|y| ()?y0 = p|y| ?)"2.)?(1) y000 = x y = 1 x4 + 1 c1x2 + c2x + c3242qy(0) = a0, y0 (0) = a1, y

3、00 (0) = a22 y = a + a x + a21 41x +x012224Rx(2) y0 = f (x) y =f (t)dt + C0qy(0) = 0R y =f (t)dtx0(3) R0 = aR R = Ceat, (a > 0)qR(0) = C = 1 R = eat, (a > 0) (4) y0 = 1 + y2, dy1+y2= dx arctan y = x + C, y = tan(x + C)qy(x0) = tan(x0 + C) = y0, C = arctan y0 x0 y = tan(x x0 + arctan y0)3.)?(1)

4、 y = Cx + x2, y0 = C + 2x y = (y0 2x)x + x2 = y0 x x2=xy0 x2 y = 0 (2)y = c1ex + c2xexy0 = c1ex + c2(xex + ex) y00 = c1ex + c2(xex + 2ex)(1a)(1b)(1c)Dy, y0 exexxexex + xex 2x= e=60= Dc1, c2 c1, c2?*d? (c1, ex(y + xy xy0 )(1a)(1b) , ex(y0 y)c2?(1c),ky00 = 2y0 y(3) ?x2 + y2 = c?>'ux? ,kx + yy0

5、= 0(4)22(x a) + (y b) = C(2a)30(x a) + (y b)y = 01 + (y0 )2 + (y b)y00 = 03y0 y00 + (y b)y000 = 0(2b)(2c)(2d)1 + (y0 )2(2c) y b = y00?(2d),k1 + (y0 )2y000 3y0 (y00 )2 = 04.y? y = g(x, c1, c2, · · · , cn)?1w?Ky? n ? y = g0 (x, c1, c2, · · · , cn) y = g00 (x, c1, c2, 

6、3; · · , cn)· · ·y = gn(x, c1, c2, · · · , cn)q c1, c2, · · · , cn*d?Dy, y0 , · · · , yn16=0Dc , c , · · · , c 1 2nd?3?n?c1, c2, · · · , cn?y, y0 , · · · , yn1L?y = gn(x, c1, c2, ·

7、 · · , cn)=? /XF (x, y, y0 , · · · , yn) = 0 ? w,?y = g(x, c1, c2, · · · , cn)?d? )? q c1, c2, · · · , cn*d?q? ?)? y"SK 1-2?41.)?(1)(2)(3)2.):(1)(2)3.)?1 ? ? ?SK 2-11.):PQ(x, y) = 0 =62 =(x, y)yx?T ?2.)?PyQ(x, y) = 2 =(x, y)x?T ?3.):PyQ(x,

8、y) = b =(x, y)x?T ?4.)?PQ(x, y) = b 6= b =(x, y), (b 6= 0)yx5?T ?5.)?PyQ(x, y) = 2t cos u =(x, y)x?T ?6.)?PQ(x, y) = ex + 2y =(x, y)yx?T ?7.)?Py1Q(x, y) =(x, y)xx?T ?8.):PyQ(x, y) = 2by,(x, y) = cyx if 2b = c, K?T ?K?T ?"9.)?P(x, y) =1 2sQ=(x, y)yt2x?T ?10.)?P0Q(x, y) = 2xyf (x2 + y2) =(x, y)yx

9、?T ?6SK 2-21.)?(1)2xy0 =, ydy = x2dxy0.5y2 = 1 x3 + C, y 6=03=3y2 = 2x3 + C,y 6= 02(2)2xxy0 =, ydy =dxy(1 + x3)1 + x310.5y2 =ln |1 + x | + C,33=3y2 2 ln |1 + x3| = C, y 6= 0, x 6= 1(3) y = 0? ?)?if y 6=0dy1 y2 = sin xdx,= cos x Cy=1 + (C + cos x)y = 0?A)y = 0(4)dyy0 = (1 + x)(1 + y2),= (1 + x)dx1 +

10、y2arctan y = 0.5x2 + x + Cy = tan(0.5x2 + x + C)(5)y0 = (cos x cos 2y)2if cos 2y 6=0dy1 + cos 2xdx= cos2 xdx =cos2 2y272 tan 2y 2x sin 2x = Cif cos 2y = 0ny =+, n Z42(6)xy0 = p1 y2y = ±1? A)?if y 6= ±1, x 6= 0dydxxp1 y2=arcsin y = ln |x| + C? )?arcsin y = ln |x| + C, (x 6= 0)and y = ±

11、1(7)0x exy =y + ey(y + ey)dy = (x ex)dx)?y2 x2 + 2(ey ex) = C (y + ey 6= 0)2.):(1)sin 2xdx + cos 3ydy = 0cos 2xsin 3y+= C23qy( ) = ,?23cos(2 )sin(3 ) 2 2 3 3+= C = 0.52 sin 3y 3 cos 2x = 38(2)xdx + yexdy = 0 xexdx + ydy = 0y2xx xe e += C2 y(0) = 1 C = e +=01 ,?122x22(x 1)e + y + 1 = 0(3)drd= rr = 0?

12、 ?)r 6= 0?drr d = 0,ln |r| = C,dr ?n?r 0,?ln r = Cqr(0) = 2, C = ln 2ln r = ln 2 r = 2e(4) ln |x| y0 =, (1 + y2)dy ln |x|dx = 01 + y2y3 y + x ln |x| + x = C3 y(1) = 0, C = 0 + 0 0 + 1 = 1y3 y + x ln |x| + x = 13(5)1 + x2y0 = xy3y = 0? ?A)?y 6= 0?xdxdy1 + x2= 0y321 + x2 += C1y29 y(0) = 1, C = 21 + 1

13、= 3+ 21 + x2 = 31y23.)?(1)y0 = cos x y = sin x + C(2)y0 = ay (a 6= 0)y = 0? ?A)?y 6= 0?dyy= adxln |y| = ax + C1y = Ceax (C 6= 0)y = 0?C = 0? A)? y = Ceax(3)y0 = 1 y2y = ±1? A)?if y 6= ±1,dy= dx1 y211 + yln | = x + C121 y=Ce2x 1y =Ce2x + 1Ce2x 1 y =and y = ±1Ce2x + 110(4)y0 = yn (n = 1

14、 , 1, 2)3if n = 1?|(2) (J?ky = Cexif n 6= 1,y = 0? ?A)?y 6= 0?yndy dx = 0 1 y1n x = C1 ny = Cexif n = 1,if n 6= 1, 1 1ny x = C and y = 01n4.)? A?B3? ?I?O?(xA, 0)?(x, y)?KdK?x xABA = p(x xA)2 + y2 = bqB $? ?A? y0 (x xA) = y?k?y0 pb2 y2 + y = 0(1)if b = 0,Ky = 0 (2)if b 6= 0, let y = b sin z, z , ?K?z

15、?2 220b(1 sin z)z + sin z = 0w,sin z = 0?=y = 0? A)?11sin z 6= 0? b 00z b sin zz + 1 = 0sin z?k11 + cos z 2 b ln 1 cos z + b cos z + x = Cp22p1b +b yb lnp+b2 y2 + x = C2b b2 y2q y(0) = b, C = 0,if b = 0,y = 01 b ln b+b2y2 pb2 y2bb2y2if b =60,x =25.y?y?w,y = a?(2.27) ?)?b L:P(x0, a)3, ? )y = g(x)?Kx1

16、, s.t. 0 < |g(x1) a| < ? g(x1) > a?-h = g(x1) a > 0?KZZa+h dyf (y) | = |x1|dx| = |x1 x0| < ax0RRRa+a+ha+ dy f (y) dy f (y) dy? = |g?| | + |aaa+h f (y)?d?3?y = a? z?:? )? "3?y = a? z?:? )? ?y = a? z?:? )?ky = aRRa±a+ dy dy f (y)b | < ,? | < af (y)aRx dy-?Kg(x) < (a x

17、 a + ),g(x) =a f (y)Kg0 (y) =6= 0 1 f (y)l d?3?ny?dgL? y = h(g)?Kdh(g)1= f (y) = f (h(g)dgg0 (y)12l y = h(x)?2.27? )w,h(x)? ua?cJg?Ra± dy b ?| y"| = af (y)6.)?p(1) f (y) =|y|3NC?Y?y = 0f (y) = 0 y = 0ZZ± dyf (y) | = |± dy|p| = 2 < |y|00?d?d?K(J?3y = 0?z?:?)?"(2) f (y) = (

18、y ln |y|,0,y 6= 0y = 03y = 0NC?Y?f (y) = 0 y =0Z± dyf (y) | = |0?d?d?K(J?3y = 0?z?:?)?"SK 2-31.)?dydx+ p(x)y = q(x) ?)?y = e R p(x)dx C + Zq(x)eR p(x)dxdx (1) ?y0 + 2y = xex ?)?y = e R 2dx C + ZxexeR 2dxdx = e2x (C + xex ex)= (x 1)ex + Ce2x(2) ?y0 + y tan x = sin(2x) ?)?y = e R tan xdx C +

19、 Z2= C cos x 2 cos xsin xeR tan xdxdx 13(3) x = 0?y = 0x 6= 0?z?dy2sin x+y =dxxx?)? Zy = e Rsin x2R2xdxdxC +edxxxC=+ (sin x x cos x)x2x2+ (0 + ) 1 =qy() =1 , = C122 C = 0sin x x cos x y =(x 6= 0)x2dy1 (4)y = 1 + x ?)?dx1x2 Z y = e R C +(1 + x)eR 11x11xdxdxdx22 !rrZ1 + x1 x=|C +(1 + x)|dx1 x1 + x y(0

20、) = 1, C = 1 3x=0 ,? ?S !rrrZ1 + x1 xy =1 +(1 + x)dx1 x1 + x1 + x 2 + arcsin x + x 1 x2(1 < x < 1)=1 x22.)?(1) -u = y2?Kdudxdydxx2 + y2= 2y= 2y2y= x2 + u14(2) ?xw?y ?K ?z?dxx + y2=dyy(3) -u = y3?Kdux= 3xy2 dydxdx33= x y3= x u(4) -u = sin y?Kdudxdy= cos ydx= cos y + x tan y 1cos y= 1 + x sin y=

21、 1 + xu3.y? y = (x)?vy0 + a(x)y 0 (x 0) 0 (x) + a(x)(x) 0 (x 0) Rx0 ea(s)ds (x) + a(x)(x) 0 (x 0)0hi0 Rxa(s)ds(x)0 (x 0)e 0RRx0 ea(s)dsa(s)ds(x) e(0) = (0) (x 0)00Rx (x)a(s)ds (0)e(x 0)04.)? ?g?5?dydx+ p(x)y = q(x)(3)k/Xy = C(x)e R p(x)dx )?K= C0 (x)e R p(x)dx C(x)p(x)e R p(x)dxdydx(4)15(3)(4) C (x)

22、e R p(x)dx = q(x)0 C(x) = Z q(x)eR p(x)dxdx + C y = e R p(x)dx C + Zq(x)eR p(x)dxdx 5.y:(1) q(x) 0, ?g?5? )?y = Ce R p(x)dxeT? ?")?Ky(x + ) = y(x)?=RRx+xp(s)dsp(s)dse= e00RRx+xp(s)ds+p(s)dse= 000Zx+p(s)ds = 0xdx ?5?Z1p¯ =p(s)ds = 00Re?Kp(x)dx = 0p¯ = 00qp(x)? ?Y?ZZZ+tp(x)dx =p(x + t)d

23、x =p(s)ds = 0 (t)00tRRt+tp(s)ds+p(s)dse= 0 (t)00RRt+t Cep(s)dsp(s)ds= Ce(t)00y(t + ) = y(t) (t, C 6= 0)? ?")?"(2) q(x)?"? )?y = e R p(x)dx C + Zq(x)eR p(x)dxdx 16e?k?)?K? ?C?s.t. ZxRRxse 0 p(s)dsC +q(s)e 0 p(t)dtds0 Zx+RRx+s= ep(s)dsq(s)e 0 p(t)dtdsC +00 ZZxx+RRRx+ssp(s)dsC +q(s)e 0 p

24、(t)dtds= C +q(s)e 0 p(t)dtdse x(5)00RRx+?qp(x), q(x)?p(s)ds =p(s)dsx0RqC3? p(s)ds?=e= 160p¯ 6= 0. p(x), q(x)?ZZx+xRRsk+q(s)e 0 p(t)dtds =q(k + )e 0p(t)dtdk (s = k + )0ZxRk= ep¯q(k)e 0 p(t)dtdk0x ZZ+xRsp¯eq(s)e 0 p(t)dtds = 00 ZZx Z+xRRssp¯eq(s)e 0 p(t)dtds =q(s)e 0 p(t)dtds000qp&

25、#175; 6= 0?l 2d(5)? RRR+xxsp¯p(t)dteq(s)eds000C =ep¯ 1RRsp(t)dtq(s)eds0 0=ep¯ 1l ? ?)? x ZZw R1Rxsy = e0 p(s)dsq(s)e 0 p(t)dtds+ep¯ 100176.y? ?y0 + y = f (x) ?)? Z xy = ex C +f (s)esds0R0-C =f (s)esds?KZxy = exf (s)esds f (x)3(, )?k.? ?(.?M ?KZ0s|C| = |f (s)e ds|Z0s|f (s)|e ds0sZ

26、 M= Me dsZxsx|y| = |f (s)eds|Zxsx|f (s)|edsxesxdsZ M= MR y = exxf (s)esds? k.)"eyk.)?b y = y1(x), y = y2(x) ?y0 + y = f (x) k.)?Ky = y1(x) y2(x)?y0 + y = 0 k.)?=y1(x) y2(x) = Cexk.?l C = 0, y1(x) = y2(x)?k.)?"18 f (x)?f (x + ) = f (x)?l ZZx+f (s)esxdsy(x + ) =xf (t + )etxdt (t = s )f (t)et

27、xdt=Zx= y(x)l dk.)?"7.y? (1) fn?H0? ?= > 0, N (), s.t. m, n N ()?k|fm fn| = 0 max |fm(x) fn(x)| < x2= > 0, N (), s.t. m, n N (), x?k|fm(x) fn(x)| < (R, | · |)? ? x, fn(x)? ?Pfn(x) f (x) (n +)e?Iy?(i)f (x)?2?(ii)|fn f | 0 (n +)'u(i)?xf (x + 2) = lim fn(x + 2)n+= lim fn(x)n+=

28、 f (x)'u(ii)?|fn f | = 0 max |fn(x) f (x)|x2= max |f (x) lim fm(x)|0nx2m+= max | lim (fn(x) fm(x) |0x2m+lim 0 max |fn(x) fm(x)|m+ x2=lim |fn(x) fm(x)|m+19 lim |fn f | =limm,n+|fn(x) fm(x)| = 0n+?d?H0? ?m" 1Rx+2(2) ? : f y =a(xs)f (s)ds?ee2a 1x f ?2?d(2.40) ? ?y?2?= : H0 H0 1e2a 1 H , P0= K,

29、 K(i) C1, C2, f , f12Zx+2ea(xs) (C1f1(s) + C2f2(s) ds(C1f1 + C2f2) = Kx Z Z x+2x+2ea(xs)f1(s)dsea(xs)f2(s)ds= C1 K+ C2 Kxx= C1(f1) + C2(f2)0(ii) f H|(f )| =Zx+2ea(xs)f (s)ds|max |K0x2xZx+2ea(xs)ds max |K| · |f | ·0x2xe2a 1= |K|f |a1=|f |a|= k|f |, (k =1)|a| y"SK 2-41.):(1) -y = ux?Ky0

30、 = u0 x + u = ?z?2uxx ?2xuxx2(2 u)du + x(1 u2)dx = 0if x 6= 0, u 6= ±12 u1du +dx = 01 u2x20? (1 + u)3 x3= C, (C 6= 0)1 ux(x + y)3 = C(x y), (C 6= 0)if u = 1 ,y = x if u = 1 , y = xx = 0? ? A)?n? ? )?(x + y)3 = C(x y) and y = x , (y 6= 2x)(2) -u = y + 2, v = x 1?K2(u 2) (v + 1) + 52u vu0 = y0 =2

31、(v + 1) (u 2) 42v u-u = zv?K?z?z2 1dz= v2 zdvif v 6= 0, z 6= ±1?z? 2 z1dzdv = 0z2 1v 3 3v(1 z) = C(1 + z) v , (c 6= 0)3v u = C(v + u) , (C 6= 0)if z = 1 ,u = v if z = 1 , u = vv = 0? A)?n? ? )?3u v = C(v + u) and u + v = 0, (2v u 6= 0)=3y x + 3 = C(x + y + 1) and x + y + 1 = 0, (2x y 4 6= 0)(3)

32、 -u = x + 2y?Ku0 = 2y0 + 1 =4u+1 ?2u1w,u = ?A)142114if u =6dx2u 1=du4u + 1u3x =ln |4u + 1| + C288x4u4u + 1 = Ce, (C 6= 0)3 u = ?1C 0? )4n? ?)?8y4x4x + 8y + 1 = Ce, (2x + 4y 1 6= 0)3(4) w,y = 0?A) 1 ?Kif y 6= 0?-u =y20y03u = 2= 2x + 2xuy3=u0 2xu = 2x3? ?5?) 2u = x2 + 1 + Cex?d2y2 = u1 = (x2 + 1 + Cex

33、 )1 and y = 02.)?(1) -u = x y?Ku0 = 1 y0 = 1 cos udu + (cos u 1)dx = 0 cos u = 1?=u = 2k, k Z?A)if cos u 6= 1du+ dx = 0cos u 1ucot+ x = C222? )?x ycot+ x = C and x y = 2k, k Z2(2) -u = wv?K?z?v3(3w + 1)dw + (4w + 2)wv2dv = 0if v 6= 0, w(4w + 2) 6= 0,3w + 1dvdw += 0w(4w + 2)v) w2(2w + 1)v4 = C, (C 6=

34、 0)w,v = 0, w(4w + 2) = 0?C 0? A)?d ? ?)?(wv)2(2wv2 + v2) = Cu2(2uv + v2) = C(3) -u = y2, v = x2?K ?z?(u + v + 3)du (4u 2v)dv = 0-w = u + 1, p = v + 2?K ?UYz?(w + p)dw (4w 2p)dp = 0-w = ph?K ?UYz?p2(h + 1)dh + p(h2 3h + 2)dp = 0C?l?) 23p = 0 and (h 1) = Cp(h 2) and h = 2 p = 0?=x2 + 2 = 0? ? )? ? )(

35、y2 x2 1)2 = C(y2 2x2 3)3 and y2 = 2x2 + 3,(y 6= 0)23(4) -u = y2, v = x2?K ?z?du2v + 3u 7=dv3v + 2u 8-w = u 1, p = v 2?K ?UYz?dw3w + 2p=dp2w + 3p-w = ph?K ?UYz?p2(2h + 3)dh + 2p(h2 1)dp = 0C?l?) h + 1 = Cp4(h 1)5 and h = 1 and p = 0 p = 0?=x2 2 = 0? ? )? ? )y2 + x2 3 = C(y2 x2 + 1)5 and y2 x2 + 1 = 0

36、3.)?(1) -u = xy?K 21 4u 4u 1 ,002u = xy + y = x(y ) + y =(x 6= 0)4x24x2if 4u 4u 1 6= 0?K?z?dudx=4u 4u2 14x) 4 2xe 12u= C, (C > 0)214u 4u 1 = 0?=u = ?A)2? ?)?12 4 2xe 12xy= C, (C > 0) and xy =(2) xy = 1?x2y0 = x2y2 + xy + 1 ?A)2400 1 -u = y + ?K1u = y ?z?xx2ux02u = u -z = u1?K?UYz?z0z + 1 = 0x)

37、 z = Cx x ln |x|? ?)?111y = x + Cx x ln |x|and y = x4.)? y = 0? ?A)if y 6= 0?-u = y1y0 ?Ku0 = y1y00 y2(y0 )2= y1 p(x)y0 q(x)y y2(y0 )22= p(x)u q(x) u=z?pk0?u0 = u2 p(x)u q(x)5.)? y = y(x)?:(x, y)?(1, y0 )?(x, y)?Y ? ,4k? 0y xy = tan( ) = 10 y 1 + y 4x y xx + yy0 =or y0 =x yx + y25e?)1?-y = ux?K?z?du

38、1 + u2x =dx1 u 12arctan u ln(1 + u ) ln |x| = C2=yarctanln(x + y ) = C122x2?u1 ?-z = y?Kz0 = x+z ?=?1?xz y122arctan( ) ln(x + y ) = Cx2=y1arctan+ln(x2 + y2) = Cx2?d?yarctan±ln(x + y ) = C122x26.)? :1 ?u :?1? ?x?1?=?y =y(x)?dA?'X?w?L?:?3? ? ?T? 2 Kk?2y0y=1 (y0 )2 x-u = y2?K?z? !20uu + x2 =+

39、x22-v = u + x2?w,v > 0?K?UYz?4v = (v0 )2012v = ±2v26 v = (x + C)2=x2 + y2 = (x + C)2? ?y2 = 2Cx + C2SK 2-51.)?(1)= 3x2 + 2x + 3y2, Q = 2xPyx 1PQx= 3Qy?fµ(x) = eR 3dx = e3x?u?ke3x(3x2y + 2xy + y3)dx + e3x(x2 + y2)dy = 0= 13x23d e (xy + y )= 0313x23e (x y + y ) = C3(2)= 1, Q = 2yPyx 1QPy1

40、y= 2 PxR1(2 )dy1 2y?fµ(y) = e?u?k= e , (y = 0)6yy 12y2y2ye dx + e2xedy = 0y27=d xe2y ln |y| = 0xe2y ln |y| = C, (y 6= 0)y = 0?A)"(3) ?xy?z?(3x2y + 6x)dx + (x3 + 3y2)dy = 0? ?kx3y + 3x2 + y3 = C(4)22(ydx xdy) (x + y )dy = 01?|k?fx2, y2, (x2 + y2)1?1 |k?f(x2 + y2)1 µ = (x2 + y2)1?kx2 + y2 + xyx2 + y2 dx dy = 0x2 + y2 yarctan+ y = C and y = 0x(5)= 6xy2, Q = 2xy2Pyx 1Q24xyPy2= y=2xy3PxR2( )dy2?fµ(y) = e?u?k= y, (y = 0)6y2xydx + (x2 y2)dy = 0 x2y + 1 = C and y = 0y(6)2(ydx xdy) + xy dx = 028w,µ = y2? ?f?z?1xdx dy + xdx = 0, (y 6= 0)yy2=x2x+= C, (y 6=