解:引进参数p=y,方程可写成参数形式 (1) (2) 为公去变量y,将(2)式对x求导后减去(1)式,得p,与x的微分方 程-p-x-p-p=0,p≠0时,将它改写成以p为自变量的线性方 程dr/dp /(2p)-1/2,它的通解是x=c(p)-1-p/3,代入(2)得 参数形式的通解x=c(±p)-1/2-p/3,y=千c(±)1/-p2/6.、当P=0 时,得特解y=0 8)x2+y2=1 解:令x=sint,t∈[-π/2,丌/2,得参数形式的微分方程 int,y/'=±cost,公去x:dy=ydr=±cos2tdt,积分得参数 形式的通解:c=simt,y=±2t+sin(2)/4+c 9)y+y3-3yy=0 解:令y=yt,代入方程得y=0或y=3/(1+t3),从前者得特 解y=0,从后者得参数方程组y=32/(1+t2),y=3/(1+t3),再由 dx=dy/y=(1+t)/(3t)dy=[-1+3/(1+t3)dt 1+1/(1+t)-(t-1/2)/(2-t+1)+3/(2(t2-t+1)dt 积分得参数形式的通解 ±(1+t) t2 y= exp(y) 引进参数p=y,方程可写成参数形式 y (p)p2 为公去变量y,将(2)式对x求导后减去(1)式,得p与x的微分方 程故(expp)p2+2pexp(P)p-p=0,整理得p(1-exp(p)(2+p)p)=0 由p=0,代入(2)式得特解y=0,由(1-exp(p)(2+p)p)=0得参数 形式的通解 6.已知f()>0在(.+∞)上微分且f(a)/f(t)dt=1,x>0,) 求f(x)的表达式 解:将方程化为/f()=fm两边对x求导得:f(x) (n,积分得P(a 代入原方程得c=0,故f(x)
: <g p = y 0 , RQHgO/ y 0 = p, (1) y = −xp − 1 2 p 2 , (2) 1.E* y, ~ (2) /Z x ]^E (1) /k v p, x −p−xp0 −pp0 −p = 0, p 6= 0 , ~6jQHS p 1Z* dx/dp = −x/(2p) − 1/2, 6 = / x = c(±p) −1/2 − p/3, 7(2)v gO/ = x = c(±p) −1/2 − p/3, y = ∓c(±p) 1/2 − p 2/6. = p = 0 k v? y = 0. 8) x 2 + y 02 = 1 : U x = sin t, t ∈ [−π/2, π/2], vgO/ x = sin t, y0 = ± cost , .E x: dy = y 0 dx = ± cos2 t dt, )vg O/ = : x = sin t, y = ±[2t + sin(2t)]/4 + c. 9) y 03 + y 3 − 3yy0 = 0 : U y = y 0 t, 7v y 0 = 0 O y 0 = 3t/(1 + t 3 ), ulmv? y = 0, umvg y = 3t 2/(1 + t 3 ), y 0 = 3t/(1 + t 3 ), dx = dy/y0 = (1 + t 3 )/(3t) dy = [−1 + 3/(1 + t 3 )] dt = [−1 + 1/(1 + t) − (t − 1/2)/(t 2 − t + 1) + 3/(2(t 2 − t + 1))] dt. )vgO/ = : x = −t + ln · ± (1 + t) √ t 2 − t + 1¸ + √ 3 arctan µ 2t − 1 √ 3 ¶ + c, y = 3t 2 1 + t 3 10) y = exp(y 0 )y 02 : <gp = y 0 , RQHgO/ y 0 = p, (1) y = exp(p)p 2 . (2) 1.E* y, ~ (2) /Z x ]^E (1) /k v p, x V(exp(p)p 2 + 2p exp(p))p 0 − p = 0, |v p(1 − exp(p)(2 + p)p 0 ) = 0, p = 0, 7 S2 /v? y = 0, (1 − exp(p)(2 + p)p 0 ) = 0 vg O/ = x = (p + 1) exp(p) + c, y = p 2 exp(p). 6. \ f(x) > 0 A(0, +∞) DFf (x) Z x 0 f (t) dt = 1, x > 0, ] f (x) 3i/. : ~+1 Z x 0 f (t) dt = 1 f (x) , wxZ x ]^v: f (x) = − f 0 (x) f 2 (x) , )v f 2 (x) = 1 2x + c , 7;v c = 0, V f (x) = 1 √ 2x . 9
7.假设x(0)程在,方求满足 (t+s) (t)+a(s 的函数x(t 令t=0,8=0,果x(0)=0,因此 lim 1+x2(t) (s [1-x(1)x(s) 1+2()im( 1+ 即x(t)满足连分试程x(t)=x(0)[1+x2(t)],积分果 arctan(x)=x(0)t+c,即r(t)=tan(x’(0)t+c),再由x(0)=0果c=0, 故果所求函数为x(t)=tan(x(0)t) 8.求一曲第,使果在正上面任一点的切第介例与标轴间的部分 被切点所初分 设在直等与标系xOy中曲第的试程为y=y(x),在点(x,y(x) 的切第与x轴,y轴组成的解等形中,由题意解等形的高角例2驯, 法角例21,因此x=-,积分果x=c,c≠0 9.设函数(t)在(-∞,+∞)上微分,x(t)不恒为零.x(0)程在 且满足条件r(t+s)=x(t)x(s),方求此函数 令t=0,s=0,果x(0)=0,或x(0)=1,之见当x(0)=0时, x(t)≡x(t)x()=0,故只考虑x(0)=1的题引 x(0)x(t), 即x()满足连分试程x()=x(0)x(t),积分果x(t)=c2(0)t, 再由x()=1果c=1,故所求函数为x(t)=e2(0)t 10.写出试程M(x,y)dx+N(x,y)dy=0一有形为u(x±y),p(xy) (x2±y2)的积分因子的充要条件 答:因为M(x,y)dx+N(x,y)dy=0一有形如μ((x,y)的积分 因子的充要条件是 My-Na Pr-Moy 常是φ的函数.所以试程有形如p(x±y),(xy),(x2y2)的积分 因子的充要条件是 M N干MNy- M:'NEFMy分别常是x±y,xy,x2±y2的函 数 11.设M(x,y),N(x,y)都是x,y的微分之连的m次齐次函数, m≠-1.记U(x,y)=xM(x,y)+yN(x,y),证系
7. w x 0 (0) A, ]HI x (t + s) = x (t) + x (s) 1 − x (t) x (s) J x (t). : Ut = 0, s = 0, v x (0) = 0, [_ x 0 (t) = lims→0 x (t + s) − x (t) s = h 1 + x 2 (t) i lims→0 x (s) s [1 − x (t) x (s)] = h 1 + x 2 (t) i lims→0 x (s) s = h 1 + x 2 (t) i x 0 (0) . x (t) HI x 0 (t) = x 0 (0) £ 1 + x 2 (t) ¤ , )v arctan(x) = x 0 (0)t + c, x (t) = tan (x 0 (0)t + c), x (0) = 0 v c = 0, Vvd]J1 x (t) = tan (x 0 (0)t). 8. ]r, evA6DUj 2zC 2jd . : Ah xOy Xr 1 y = y (x), Aj(x, y (x)) 2x , y H OX, V O z 2|y|, z 2|x|, [_ x dy dx = −y, )v xy = c, c 6= 0. 9. J x (t) A (−∞, +∞) D, x (t) YG1$. x 0 (0)A, FHIst x (t + s) = x (t) x (s), ]_J. : U t = 0, s = 0, v x(0) = 0, O x(0) = 1, RF= x(0) = 0 , x(t) ≡ x(t)x(0) ≡ 0, V x(0) = 1 , x 0 (t) = lims→0 x (t + s) − x (t) s = x (t) lims→0 x (s) − 1 s = x 0 (0) x (t), x (t) HI x 0 (t) = x 0 (0) x (t), )v x (t) = ce x 0 (0)t , x (0) = 1 v c = 1, Vd]J1 x (t) = ex 0 (0)t . 10.Q) M (x, y) dx+N (x, y) dy = 0 8O1 µ (x ± y), µ (xy), µ ¡ x 2 ± y 2 ¢ )[d Qst. 2: [1 M (x, y) dx + N (x, y) dy = 0 8Op µ (ϕ (x, y)) ) [d Qst/: My − Nx Nφx − Mφy / ϕ J. dS8Op µ (x ± y), µ (xy), µ ¡ x 2 ± y 2 ¢ ) [d Qst/ My − Nx N ∓ M , My − Nx Ny − Mx , My − Nx Nx ∓ My , :/ x ± y, xy, x 2 ± y 2 J . 11. M (x, y), N (x, y) $/ x, y R m 101J, m 6= −1. ^U (x, y) = xM (x, y) + yN (x, y), p 10
1)aM+yMy=mM(a,y), IN2+yNy=M 2)若M(x,y)dx+N(xr,y)dy=0为全微分方程,则其通解为 ,y)= 3)(x,y) U(a, y) 是方程M(x,y)dx+N(x,y)dy=0的积分因 子 证:1)因M(x,y)是x,y的m次齐次函数,即对于任何t>0,成 立恒等式M(tx,ty)≡tmM(xr,y).由M的可微性,两边对t求导,得 rM1(tx,ty)+yM2(tr,ty)≡mtm-1M(x,y),其中M(tx,ty)、M2(tr,ty) 分别表示函数对第一、第二个自变量求偏导数令上式中t=1即 得证.同理可证关于N(x,y)的恒等式 2)因m≠-1,原微分方程等价于(1+m)(Mdx+Ndy)=0, 又因Mdr+Ndy=0为全微分方程,所以My=Nx dl du= M dr+Ndy+(aMr +y Nr)dc+(aM,+y Ny)dy 又因Mdx+Ndy=0为全微分方程,有M=Nx,所以 (a, y)dr M(, y) d 再因M(x,y),N(x,y)都是x,y的m次齐次函数,所以由1)得 dU=(1+m)(Mdx+Ndy),即U(x,y)=c是方程的通解 3)由积分因子的定义,只要证明2(AM-0M)=0即可 求导得 0(M)O(N)1 =r[U (My-N)+NUr-MUJI 其中 Ux=M+Mr+yNr, Uy=N+aMy+yNy, (3) 将(3)代入(2)整理得 0(M)a(N)_1 2IN(Mr+yMy)-M(aN+y Ny),(4) 再将(1)代入(4)得 0(uM)(N)=0.证毕 12.已知下列 Riccati方程的一个特解φ(x),试求出其通解 1)ye-I+y2-2ye=1-e2r, (a) 解:令y=ex+-1,代入方程得未知函数u的方程n′=e2,积分 得u=e2+c,因此原方程的通解为y=e+(er+c)-1 2)y+y2-2y sin r= cos -sin2, p(a) 解:令y=sinx+a-1,代入方程得未知函数u的方程v=1,积 分得w=x+c,因此原方程的通解为y=sinx+(x+c)-1 解:令y=-(2x)-1+-1,代入方程得未知函数u的方程 u=x-1a-1,积分得u=-xln(cx),因此原方程的通解为 (2x)-1-(xln(cr)-1
1) xMx + yMy ≡ mM (x, y), xNx + yNy ≡ mN, 2) M (x, y) dx + N (x, y) dy = 0 1], aq= 1 U (x, y) = c. 3)µ (x, y) = 1 U (x, y) / M (x, y) dx + N (x, y) dy = 0 )[ d. p: 1) [ M (x, y) / x, y m 101J, ZzU; t > 0, H bG / M (tx, ty) ≡ t mM (x, y). M R, wxZ t ]^, v xM0 1 (tx, ty)+yM0 2 (tx, ty) ≡ mtm−1M (x, y), qX M0 1 (tx, ty)> M0 2 (tx, ty) :34JZ> NZ*]^.UD/X t = 1 vp. Rpyz N (x, y) G /. 2) [ m 6= −1, ; z (1 + m) (M dx + N dy) = 0, [ M dx + N dy = 0 1], dS My = Nx, a dU = M dx + N dy + (xMx + yNx) dx + (xMy + yNy) dy, [ M dx + N dy = 0 1], 8 My = Nx, dS dU = M (x, y) dx + M (x, y) dy + (xMx + yMy) dx + (xNx + yNy) dy, [ M (x, y), N (x, y) $/ x, y m 101J, dS1)v dU = (1 + m) (M dx + N dy), U (x, y) = c / = . 3))[d !E, Qp ∂ (µM) ∂y − ∂ (µN) ∂x = 0 R. ]^v ∂ (µM) ∂y − ∂ (µN) ∂x = 1 U2 [U (My − Nx) + NUx − MUy] , (2) qX Ux = M + xMx + yNx, Uy = N + xMy + yNy, (3) ~(3)7(2)|v ∂ (µM) ∂y − ∂ (µN) ∂x = 1 U2 [N (xMx + yMy) − M (xNx + yNy)] , (4) ~(1)7(4)v ∂ (µM) ∂y − ∂ (µN) ∂x = 0. p . 12. \*+Riccati N? ϕ(x), ])q= : 1) y 0 e −x + y 2 − 2ye x = 1 − e 2x , ϕ(x) = ex : : U y = ex + u −1 , 7v[\J u u 0 = ex , ) v u = ex + c, [_; = 1 y = ex + (ex + c) −1 . 2) y 0 + y 2 − 2y sin x = cos x − sin2 x, ϕ(x) = sin x : : U y = sin x + u −1 , 7v[\J u u 0 = 1, ) v u = x + c, [_; = 1 y = sin x + (x + c) −1 . 3) 4x 2 (y 0 − y 2 ) = 1, ϕ(x) = − 1 2x : : U y = −(2x) −1 + u −1 , 7v[\J u u 0 = x −1u − 1, )v u = −x ln(cx), [_; = 1 y = −(2x) −1 − (x ln(cx))−1 . 11
4)x2y/+(xy-2)2=0,y(x) 1 解:令y=x-1+-1,代入方程得未知函数a的方程u=-x-1a+1, 积分得=x-1(+x2/2,因此原方程的通解为y=x-1+r(c+x2/2)-1. 5)y=(x-1)y2+(1-2x)y+x,p(x)= 解:令y=1+x-1,代入方程得未知函数u的方程v=+1-x 积分得u=e(c+re-x),因此原方程的通解为y=1+(ce2+x) 13.求出 logistic方程的初值问题 r(1--),x(0)=x0≠0,(>0,x是常数) 解x(t),并算出.limx(t)的值和说明其实际意义 解:这是n=2时的 Bernoulli方程,除了平凡解x≡0外,其解可 写成形式 rt) 故可见,limr(t)=xf,说明了只要x(0)≠0,任何解x趋向平衡态 14*.求出初值问题 d=8f()-k,k(O)=k0 的解k(t,s),其中f(k)=ak8为Cob- Douglas生产函数,这里a>0 0<B<1均为常数:算出kx(s)=imk(t,s)和时间无限增大时的 人均消费c∞(s)=(1-s)∫(k(s),由此求出使得cx(s)达到最大值 的s值,从而验证资本积累的黄金准则f(k(s)=p的正确性 解:这是n=B时的 bernoulli方程,除了无实际意义的平凡解 k≡0外为 k-(,s)=b-c(-+3(1 -(1-)t 所以 (s)= lim k(t, s) e∞()=(1-s)f(kx()=a(1-s)(2)F 求它的最大值可得,当s=B时取最大值 容易验证资本积累的黄金准则f(kx(s)=μ的正确性 15*在习题1.1的第九题中,当赛艇从静止状态开始运动时,求出 其运动速度u与时间t的关系,并说明其实际意义
4) x 2 y 0 + (xy − 2)2 = 0, ϕ(x) = 1 x : : U y = x −1+u −1 , 7v[\J u u 0 = −x −1u+1, )v u = x −1 (c+x 2/2), [_; = 1 y = x −1+x(c+x 2/2)−1 . 5) y 0 = (x − 1)y 2 + (1 − 2x)y + x, ϕ(x) = 1. : U y = 1 + u −1 , 7v[\J u u 0 = u + 1 − x, )v u = ex (c + xe −x ), [_; = 1 y = 1 + (ce x + x) −1 . 13. ])logistic @ dx dt = rx(1 − x xf ), x(0) = x0 6= 0, (r > 0, xf/) x(t),,`) lim t→+∞ x(t) @' q VE. : f/ n = 2 Bernoulli , I ! x ≡ 0 ", q R QHO/ x(t) = xf 1 − µ 1 − xf x0 ¶ e−rt VRF lim t→+∞ x(t) = xf , Q x(0) 6= 0, U; x #$ %& x = xf . 14*. ]) @ dk dt = sf(k) − µk, k(0) = k0 k(t, s), qX f(k) = akβ 1 Cobb-Douglas '(J, f a > 0, 0 < β < 1 %1: `) k∞(s) = limt→∞ k(t, s) ' C)*+, -%./ c∞(s) = (1 − s)f(k∞(s)), _])ev c∞(s) ii0,@ s @, uWop1)2 345a f 0 (k∞(s)) = µ 66. : f/ n = β Bernoulli , I) VE ! k ≡ 0 "1 k 1−β (t, s) = k 1−β 0 e −µ(1−β)t + as µ ³ 1 − e −µ(1−β)t ´ . dS k∞(s) = limt→∞ k(t, s) = µ as µ ¶ 1 1−β , c∞(s) = (1 − s)f(k∞(s)) = a(1 − s) µ as µ ¶ β 1−β . ]6 0,@Rv, = s = β k0,@ max c∞(s) = a(1−β) ³ aβ µ ´ β 1−β , -op1)2 345af 0 (k∞(s)) = µ 66. 15* As1.1 .X,= u/01&2r_a ,]) q_ah" u C t yk , q VE. 12
解:记a=(P/k)1/3,则方程可化为 积分并利用初始条件v(0)=0,得解为 u2+ 2u+a、6kat 2√3 arctan( 可见,当t→+∞时,u→a.说明了最后随着速度的增加,阻力 增大,牵引力减少,最终趋于平衡,速度趋于一个定值a
N ^ a = (p/k) 1/3 , aR+1 [− 2 u − a + (2u + a) − 3a u 2 + au + a 2 ]du = 6ka m dt. ),X9 rst u(0) = 0, v 1 ln u 2 + au + a 2 (u − a) 2 + π √ 3 − 2 √ 3 arctan(2u + a √ 3a ) = 6kat m . RFk = t → +∞ k u → a. 0 #h" +ok %k +,k $kk 0P#z %k h"#zN!@ a. 13