第一二章

1 1˜Ù Ä�Vg SK 1-1 1.)µ (1) ∵ y = C1e2x + C2e−2x ∴ y′ = 2C1e2x − 2C2e−2x y ′′ = 4C1e 2x + 4C2e −2x ∴ y′′ − 4y = 0 q∵ D[y,y ′ ] D[C1,C2] = ∣∣∣∣∣ e2x e−2x2e2x −2e−2x ∣∣∣∣∣ = −4 6= 0 ∴ d¼ê´mýƒA‡©§�Ï)" (2) ∵ y = sin x x ∴ y′ = x cos x−sin x x2 xy ′ + y = cosx ∴ y = sin x x ´xy ′ + y = cosx�˜‡A)" (3) ∵ y = x( ∫ x−1exdx+ C) y ′ = ∫ x−1exdx+ C + ex ∴ xy′ − y = xex q∵ dy dc = x 6= 0, (x 6= 0) ∴ y = x( ∫ x−1exdx+ C)´xy ′ − y = xex�Ï)" (4) ∵ y =  − 1 4 (x− c1)2 ,−∞ < x < c1 0 , c1 ≤ x ≤ c2 1 4 (x− c2)2 , c2 < x <∞ (∗) ∴ y′ =  − 1 2 (x− c1) ,−∞ < x < c1 0 , c1 ≤ x ≤ c2 1 2 (x− c2) , c2 < x <∞ y ′ = √|y| ∴ (∗)´y′ = √|y|�Ï)" 2.)µ (1) ∵ y′′′ = x ∴ y = 1 24 x4 + 1 2 c1x 2 + c2x+ c3 qy(0) = a0, y ′ (0) = a1, y ′′ (0) = a2 2 ∴ y = a0 + a1x+ 12a2x2 + 1 24 x4 (2) y ′ = f(x) ⇒ y = ∫ x 0 f(t)dt+ C qy(0) = 0 ∴ y = ∫ x 0 f(t)dt (3) R ′ = −aR ⇒ R = Ce−at, (a > 0) qR(0) = C = 1 ∴ R = e−at, (a > 0) (4) ∵ y′ = 1 + y2, ∴ dy 1+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.)µ (1) y = Cx+ x2, ∴ y′ = C + 2x ∴ y = (y′ − 2x)x+ x2 = y′x− x2 =xy ′ − x2 − y = 0 (2) y = c1e x + c2xe x (1a) y ′ = c1e x + c2(xe x + ex) (1b) y ′′ = c1e x + c2(xe x + 2ex) (1c) D[y, y ′ ] D[c1, c2] = ∣∣∣∣∣ ex xexex ex + xex ∣∣∣∣∣ = e2x 6= 0 ∴ c1, c2´*dÕá� (1a)(1b) ⇒ { c1 , e −x(y + xy − xy′) c2 , e −x(y ′ − y) “\(1c),k y ′′ = 2y ′ − y (3) éx2 + y2 = cü>'ux¦�,k x+ yy ′ = 0 (4) (x− a)2 + (y − b)2 = C (2a) 3 (x− a) + (y − b)y′ = 0 (2b) 1 + (y ′ )2 + (y − b)y′′ = 0 (2c) 3y ′ y ′′ + (y − b)y′′′ = 0 (2d) (2c)⇒ y − b = −1 + (y ′ )2 y′′ “\(2d),k [1 + (y ′ )2]y ′′′ − 3y′(y′′)2 = 0 4.yµ ∵ y = g(x, c1, c2, · · · , cn)¿©1w§Ky†�n�Œ� y = g ′ (x, c1, c2, · · · , cn) y = g ′′ (x, c1, c2, · · · , cn) · · · y = gn(x, c1, c2, · · · , cn) q∵ c1, c2, · · · , cn*dÕ᧠∴ D[y, y ′ , · · · , yn−1] D[c1, c2, · · · , cn] 6= 0 dÛ¼ê3½n§c1, c2, · · · , cnŒ±^y, y′ , · · · , yn−1Lѧ“\ § y = gn(x, c1, c2, · · · , cn) =Œ�/XF (x, y, y ′ , · · · , yn) = 0�§ w,§y = g(x, c1, c2, · · · , cn)´d§�)§ q∵ c1, c2, · · · , cn*dÕ᧠∴q´§�Ï)§�y" SK 1-2 4 1.)µ (1) (2) (3) 2.): (1) (2) 3.)µ 1�Ù Ð�È©{ SK 2-1 1.): ∵ ∂P ∂y (x, y) = 0 6= 2 = ∂Q ∂x (x, y) ∴Ø´T�§ 2.)µ ∵ ∂P ∂y (x, y) = 2 = ∂Q ∂x (x, y) ∴´T�§ 3.): ∵ ∂P ∂y (x, y) = b = ∂Q ∂x (x, y) ∴´T�§ 4.)µ ∵ ∂P ∂y (x, y) = −b 6= b = ∂Q ∂x (x, y), (b 6= 0) 5 ∴Ø´T�§ 5.)µ ∵ ∂P ∂y (x, y) = 2t cosu = ∂Q ∂x (x, y) ∴´T�§ 6.)µ ∵ ∂P ∂y (x, y) = ex + 2y = ∂Q ∂x (x, y) ∴´T�§ 7.)µ ∵ ∂P ∂y (x, y) = 1 x = ∂Q ∂x (x, y) ∴´T�§ 8.): ∵ ∂P ∂y (x, y) = 2by, ∂Q ∂x (x, y) = cy ∴ if 2b = c, K´T�§¶ÄK§Ø´T�§" 9.)µ ∵ ∂P ∂y (x, y) = 1− 2s t2 = ∂Q ∂x (x, y) ∴´T�§ 10.)µ ∵ ∂P ∂y (x, y) = 2xyf ′ (x2 + y2) = ∂Q ∂x (x, y) ∴´T�§ 6 SK 2-2 1.)µ (1) y ′ = x2 y , ydy = x2dx 0.5y2 = 1 3 x3 + C, y 6= 0 = 3y2 = 2x3 + C, y 6= 0 (2) y ′ = x2 y(1 + x3) , ydy = x2 1 + x3 dx 0.5y2 = 1 3 ln |1 + x3|+ C, = 3y2 − 2 ln |1 + x3| = C, y 6= 0, x 6= −1 (3) y = 0´§�˜‡)§ if y 6= 0 −dy y2 = sinxdx, 1 y = − cosx− C = 1 + (C + cosx)y = 0 ÚA) y = 0 (4) y ′ = (1 + x)(1 + y2), dy 1 + y2 = (1 + x)dx arctan y = 0.5x2 + x+ C y = tan(0.5x2 + x+ C) (5) y ′ = (cosx cos 2y)2 if cos 2y 6= 0 dy cos2 2y = cos2 xdx = 1 + cos 2x 2 dx 7 2 tan 2y − 2x− sin 2x = C if cos 2y = 0 y = pi 4 + npi 2 , n ∈ Z (6) xy ′ = √ 1− y2 y = ±1´§�A)§ if y 6= ±1, x 6= 0 dy√ 1− y2 = dx x arcsin y = ln |x|+ C ∴§�) arcsin y = ln |x|+ C, (x 6= 0)and y = ±1 (7) y ′ = x− e−x y + ey (y + ey)dy = (x− e−x)dx ) y2 − x2 + 2(ey − e−x) = C (y + ey 6= 0) 2.): (1) sin 2xdx+ cos 3ydy = 0 −cos 2x 2 + sin 3y 3 = C qy(pi 2 ) = pi 3 ,¤± −cos(2 ∗ pi 2 ) 2 + sin(3 ∗ pi 3 ) 3 = C = 0.5 ∴ 2 sin 3y − 3 cos 2x = 3 8 (2) xdx+ ye−xdy = 0⇒ xexdx+ ydy = 0 ⇒ xex − ex + y 2 2 = C ∵ y(0) = 1 ∴ C = −e0 + 1 2 = − 1 2 ,¤± 2(x− 1)ex + y2 + 1 = 0 (3) dr dθ = r r = 0´§�˜‡) r 6= 0ž§ dr r − dθ = 0, ln |r| − θ = C, dr�Ôn¿Âr ≥ 0,¤±ln r − θ = C qr(0) = 2, ∴ C = ln 2 ln r − θ = ln 2⇒ r = 2eθ (4) ∵ y′ = ln |x| 1 + y2 , (1 + y2)dy − ln |x|dx = 0 ∴ y + y 3 3 − x ln |x|+ x = C ∵ y(1) = 0,∴ C = 0 + 0− 0 + 1 = 1 ∴ y + y 3 3 − x ln |x|+ x = 1 (5) √ 1 + x2y ′ = xy3 y = 0´§�˜‡A)§ y 6= 0ž§ xdx√ 1 + x2 − dy y3 = 0 2 √ 1 + x2 + 1 y2 = C 9 ∵ y(0) = 1, ∴ C = 2 √ 1 + 1 = 3 ∴ 1 y2 + 2 √ 1 + x2 = 3 3.)µ (1) y ′ = cosx ∴ y = sinx+ C (2) y ′ = ay (a 6= 0) y = 0´§�˜‡A)§ y 6= 0ž§ dy y = adx ln |y| = ax+ C1 y = Ceax (C 6= 0) y = 0´C = 0ž�A)§� y = Ceax (3) y ′ = 1− y2 y = ±1´§�A)§ if y 6= ±1, dy 1− y2 = dx 1 2 ln |1 + y 1− y | = x+ C1 = y = Ce2x − 1 Ce2x + 1 ∴ y = Ce 2x − 1 Ce2x + 1 and y = ±1 10 (4) y ′ = yn (n = 1 3 , 1, 2) if n = 1§|^(2)�(J§k y = Cex if n 6= 1, y = 0´§�˜‡A)§ y 6= 0ž§ y−ndy − dx = 0 1 1− ny 1−n − x = C ∴ if n = 1, y = Ce x if n 6= 1, 1 1−ny 1−n − x = C and y = 0 4.)µ �A§B3Әž�‹I©O(xA, 0)Ú(x, y)§KdK¿ x ≤ xA BA = √ (x− xA)2 + y2 = b qB�$Е[�•A§� y ′ (x− xA) = y üªéá§k‡©§ y ′√ b2 − y2 + y = 0 (1)if b = 0,K y = 0 (2)if b 6= 0, let y = b sin z, z ∈ [−pi 2 , pi 2 ]§K‡©§Œz b(1− sin2 z)z′ + sin z = 0 w,sin z = 0§=y = 0´‡©§�A)§ 11 sin z 6= 0ž§ b sin z z ′ − b sin zz′ + 1 = 0 È©§k −1 2 b ln 1 + cos z 1− cos z + b cos z + x = C −1 2 b ln b+ √ b2 − y2 b−√b2 − y2 +√b2 − y2 + x = C q∵ y(0) = b, ∴ C = 0, ∴ if b = 0, y = 0 if b 6= 0, x = 1 2 b ln b+ √ b2−y2 b− √ b2−y2 − √ b2 − y2 5.yµ ⇐ £‡y¤w,y = a´§(2.27)�˜‡)§ b�L:P(x0, a)3, ˜‡ØÓ�)y = g(x)§K ∃x1, s.t. 0 < |g(x1)− a| < ε ؔ�g(x1) > a§-h = g(x1)− a > 0§K | ∫ a+h a dy f(y) | = | ∫ x1 x0 dx| = |x1 − x0| <∞ ù†∞ = | ∫ a+ε a dy f(y) | ≤ | ∫ a+h a dy f(y) |+ | ∫ a+ε a+h dy f(y) |gñœ Ïd§3†‚y = aþ�z˜:§§�)´Û܍˜�" ⇒ ∵3†‚y = aþ�z˜:§§�)´Û܍˜� ∴†‚y = aþ�z˜:§§�)ky = a b�| ∫ a±ε a dy f(y) | <∞,ؔ�| ∫ a+ε a dy f(y) | <∞ -g(x) = ∫ x a dy f(y) §Kg(x) <∞ (a ≤ x ≤ a+ ε), Kg ′ (y) = 1 f(y) 6= 0 l dÛ¼ê3½nyŒdgLѧؔ�y = h(g)§K dh(g) dg = 1 g′(y) = f(y) = f(h(g)) 12 l y = h(x)´§£2.27¤�) w,h(x)Øð�ua§ù†cJgñœ �b�ؤá§| ∫ a±ε a dy f(y) | =∞ �y" 6.)µ (1) f(y) = √|y|3y = 0NCëY§…f(y) = 0⇐⇒ y = 0 | ∫ ±ε 0 dy f(y) | = | ∫ ±ε 0 dy√|y| | = 2√ε <∞ Ïd§dþK(J§3y = 0þz˜:§§)؍˜" (2) f(y) = { y ln |y|, y 6= 0 0, y = 0 3y = 0NCëY§…f(y) = 0 ⇐⇒ y = 0 | ∫ ±ε 0 dy f(y) | =∞ Ïd§dþK(J§3y = 0þz˜:§§)Û܍˜" SK 2-3 1.)µ ® dy dx + p(x)y = q(x) �Ï)´ y = e− ∫ p(x)dx ( C + ∫ q(x)e ∫ p(x)dxdx ) (1) §y ′ + 2y = xe−x�Ï) y = e− ∫ 2dx ( C + ∫ xe−xe ∫ 2dxdx ) = e−2x (C + xex − ex) = (x− 1)e−x + Ce−2x (2) §y ′ + y tanx = sin(2x)�Ï) y = e− ∫ tan xdx ( C + ∫ sinxe ∫ tan xdxdx ) = C cosx− 2 cos2 x 13 (3) x = 0ž§y = 0 x 6= 0ž§§Œz dy dx + 2 x y = sinx x Ï) y = e− ∫ 2 xdx ( C + ∫ sinx x e ∫ 2 xdxdx ) = C x2 + (sinx− x cosx)x−2 qy(pi) = 1 pi , = C pi2 + (0 + pi) 1 pi2 = 1 pi ∴ C = 0 ∴ y = sinx− x cosx x2 (x 6= 0) (4) dy dx − 1 1−x2 y = 1 + x�Ï) y = e − ∫ − 1 1−x2 dx ( C + ∫ (1 + x)e ∫ − 1 1−x2 dxdx ) = √ |1 + x 1− x | ( C + ∫ (1 + x) √ |1− x 1 + x |dx ) ∵ y(0) = 1, ∴ C = 1 ∴ 3x=0�,‡�S y = √ 1 + x 1− x ( 1 + ∫ (1 + x) √ 1− x 1 + x dx ) = √ 1 + x 1− x 2 + arcsinx+ x √ 1− x2 2 (−1 < x < 1) 2.)µ (1) -u = y2§K du dx = 2y dy dx = 2y x2 + y2 2y = x2 + u 14 (2) òxw‰y�¼ê§K�§z dx dy = x+ y2 y (3) -u = y3§K x du dx = 3xy2 dy dx = −x3 − y3 = −x3 − u (4) -u = sin y§K du dx = cos y dy dx = cos y ( 1 cos y + x tan y ) = 1 + x sin y = 1 + xu 3.yµ ∵ y = ϕ(x)÷vy′ + a(x)y ≤ 0 (x ≥ 0) ∴ ϕ′(x) + a(x)ϕ(x) ≤ 0 (x ≥ 0) ∴ e ∫ x 0 a(s)ds ( ϕ ′ (x) + a(x)ϕ(x) ) ≤ 0 (x ≥ 0) ∴ [ e ∫ x 0 a(s)dsϕ(x) ]′ ≤ 0 (x ≥ 0) ∴ e ∫ x 0 a(s)dsϕ(x) ≤ e ∫ 0 0 a(s)dsϕ(0) = ϕ(0) (x ≥ 0) ∴ ϕ(x) ≤ ϕ(0)e− ∫ x 0 a(s)ds (x ≥ 0) 4.)µ �šàg‚5§ dy dx + p(x)y = q(x) (3) k/Xy = C(x)e− ∫ p(x)dx�)§K dy dx = C ′ (x)e− ∫ p(x)dx − C(x)p(x)e− ∫ p(x)dx (4) 15 (3)(4)⇒ C ′(x)e− ∫ p(x)dx = q(x) ∴ C(x) = ∫ q(x)e ∫ p(x)dxdx+ C ∴ y = e− ∫ p(x)dx ( C + ∫ q(x)e ∫ p(x)dxdx ) 5.y: (1) ∵ q(x) ≡ 0, ∴§´àg‚5‡©§§ §�)y = Ce− ∫ p(x)dx ⇒ eT§�?˜š")±ω±Ï§Ky(x+ ω) = y(x)§= e− ∫ x+ω 0 p(s)ds = e− ∫ x 0 p(s)ds e− ∫ x+ω 0 p(s)ds+ ∫ x 0 p(s)ds = 0∫ x+ω x p(s)ds = 0 dx�?¿5§ p¯ = 1 ω ∫ ω 0 p(s)ds = 0 ⇐ ep¯ = 0§K ∫ ω 0 p(x)dx = 0 qp(x)±ω±Ï�ëY¼ê§¤±∫ ω 0 p(x)dx = ∫ ω 0 p(x+ t)dx = ∫ ω+t t p(s)ds = 0 (∀t) e− ∫ t+ω 0 p(s)ds+ ∫ t 0 p(s)ds = 0 (∀t) ∴ Ce− ∫ t+ω 0 p(s)ds = Ce− ∫ t 0 p(s)ds (∀t) y(t+ ω) = y(t) (∀t, C 6= 0) ∴§�?˜š")±ω±Ï" (2) q(x)Øð"§§�) y = e− ∫ p(x)dx ( C + ∫ q(x)e ∫ p(x)dxdx ) ⇒ 16 e§k˜ω±Ï)§K∃˜�~êC§s.t. e− ∫ x 0 p(s)ds ( C + ∫ x 0 q(s)e ∫ s 0 p(t)dtds ) = e− ∫ x+ω 0 p(s)ds ( C + ∫ x+ω 0 q(s)e ∫ s 0 p(t)dtds ) e ∫ x+ω x p(s)ds ( C + ∫ x 0 q(s)e ∫ s 0 p(t)dtds ) = C + ∫ x+ω 0 q(s)e ∫ s 0 p(t)dtds (5) qp(x), q(x)±ω±Ï§∴ ∫ x+ω x p(s)ds = ∫ ω 0 p(s)ds qC3…˜§�e ∫ ω 0 p(s)ds 6= 1§= p¯ 6= 0. ⇐ ∵ p(x), q(x)±ω±Ï§ ∴ ∫ x+ω ω q(s)e ∫ s 0 p(t)dtds = ∫ x 0 q(k + ω)e ∫ k+ω 0 p(t)dtdk (s = k + ω) = ep¯ω ∫ x 0 q(k)e ∫ k 0 p(t)dtdk ∴ (∫ ω+x ω −ep¯ω ∫ x 0 ) q(s)e ∫ s 0 p(t)dtds = 0 ∴ (∫ ω+x 0 −ep¯ω ∫ x 0 ) q(s)e ∫ s 0 p(t)dtds = ∫ ω 0 q(s)e ∫ s 0 p(t)dtds qp¯ 6= 0§l 2d(5)Œ� C = (∫ ω+x 0 −ep¯ω ∫ x 0 ) q(s)e ∫ s 0 p(t)dtds ep¯ω − 1 = ∫ ω 0 q(s)e ∫ s 0 p(t)dtds ep¯ω − 1 l §�ω±Ï) y = e− ∫ x 0 p(s)ds ( 1 ep¯ω − 1 ∫ w 0 + ∫ x 0 ) q(s)e ∫ s 0 p(t)dtds 17 6.yµ §y ′ + y = f(x)�Ï) y = e−x ( C + ∫ x 0 f(s)esds ) -C = ∫ 0 −∞ f(s)e sds§K y = e−x ∫ x −∞ f(s)esds ∵ f(x)3(−∞,∞)þk.§Ø”�Ùþ(.M§K |C| = | ∫ 0 −∞ f(s)esds| ≤ ∫ 0 −∞ |f(s)|esds ≤M ∫ 0 −∞ esds = M |y| = | ∫ x −∞ f(s)es−xds| ≤ ∫ x −∞ |f(s)|es−xds ≤M ∫ x −∞ es−xds = M ∴ y = e−x ∫ x −∞ f(s)e sds´§�k.)" eyk.)Ž˜µ b�y = y1(x), y = y2(x)�´§y ′ + y = f(x)�k.)§ Ky = y1(x)− y2(x)´§y′ + y = 0�k.)§ =y1(x)− y2(x) = Ce−xk.§l C = 0, y1(x) = y2(x)§k.)Ž˜" 18 �f(x)±ω±Ïž§f(x+ ω) = f(x)§l y(x+ ω) = ∫ x+ω −∞ f(s)es−x−ωds = ∫ x −∞ f(t+ ω)et−xdt (t = s− ω) = ∫ x −∞ f(t)et−xdt = y(x) l dk.)±ω±Ï" 7.yµ (1) �{fn}´H0¥˜Ä�§= ∀ε > 0,∃N(ε), s.t. ∀m,n ≥ N(ε)§k ||fm − fn|| = max 0≤x≤2pi |fm(x)− fn(x)| < ε =∀ε > 0,∃N(ε), s.t. ∀m,n ≥ N(ε),∀x§k|fm(x)− fn(x)| < ε ∵ (R, | · |)´���§ ∴ ∀x, {fn(x)}´Âñ�§Pfn(x)→ f(x) (n→ +∞) e¡Iyµ(i)f(x)±2pi±Ï (ii)||fn − f || → 0 (n→ +∞) 'u(i)§∀x f(x+ 2pi) = lim n→+∞ fn(x+ 2pi) = lim n→+∞ fn(x) = f(x) 'u(ii)§ ||fn − f || = max 0≤x≤2pi |fn(x)− f(x)| = max 0≤x≤2pi |fn(x)− lim m→+∞ fm(x)| = max 0≤x≤2pi | lim m→+∞ (fn(x)− fm(x)) | ≤ lim m→+∞ max 0≤x≤2pi |fn(x)− fm(x)| = lim m→+∞ ||fn(x)− fm(x)|| 19 ∴ lim n→+∞ ||fn − f || = lim m,n→+∞ ||fn(x)− fm(x)|| = 0 Ïd§H0´��˜m" (2) ½Âϕ : f 7−→ y = 1 e2api−1 ∫ x+2pi x e−a(x−s)f(s)ds§ ∵ f±2pi±Ï§d(2.40)�í�§y±2pi±Ï = ϕ : H0 → H0 (i) ∀C1, C2, f1, f2 ∈ H0, P 1e2api−1 = K, K ϕ(C1f1 + C2f2) = K ∫ x+2pi x e−a(x−s) (C1f1(s) + C2f2(s)) ds = C1 [ K ∫ x+2pi x e−a(x−s)f1(s)ds ] + C2 [ K ∫ x+2pi x e−a(x−s)f2(s)ds ] = C1ϕ(f1) + C2ϕ(f2) (ii) ∀f ∈ H0 ||ϕ(f)|| = max 0≤x≤2pi |K ∫ x+2pi x e−a(x−s)f(s)ds| ≤ max 0≤x≤2pi |K| · ||f || · ∫ x+2pi x e−a(x−s)ds = |K|e 2api − 1 a ||f || = 1 |a| ||f || = k||f ||, (k = 1|a|) �y" SK 2-4 1.): (1) -y = ux§Ky ′ = u ′ x+ u = 2ux−x 2x−ux§ �§Œzµ x2(2− u)du+ x(1− u2)dx = 0 if x 6= 0, u 6= ±1 2− u 1− u2 du+ 1 x dx = 0 20 È©§Œ� (1 + u)3 1− u x3 x = C, (C 6= 0) (x+ y)3 = C(x− y), (C 6= 0) if u = 1 , y = x if u = −1 , y = −x x = 0Ø´�§�A)§nþ§�§�) (x+ y)3 = C(x− y) and y = x , (y 6= 2x) (2) -u = y + 2, v = x− 1§K u ′ = y ′ = 2(u− 2)− (v + 1) + 5 2(v + 1)− (u− 2)− 4 = 2u− v 2v − u -u = zv§K§?˜Úz z2 − 1 2− z = v dz dv if v 6= 0, z 6= ±1§§z 2− z z2 − 1dz − 1 v dv = 0 � v(1− z) = C(1 + z)3v3, (c 6= 0) v − u = C(v + u)3, (C 6= 0) if z = 1 , u = v if z = −1 , u = −v v = 0Ø´§�A)§nþ§�§�) u− v = C(v + u)3 and u+ v = 0, (2v − u 6= 0) = y − x+ 3 = C(x+ y + 1)3 and x+ y + 1 = 0, (2x− y − 4 6= 0) (3) -u = x+ 2y§Ku ′ = 2y ′ + 1 = 4u+1 2u−1§ w,u = − 1 4 ´A) 21 if u 6= − 1 4 dx du = 2u− 1 4u+ 1 x = u 2 − 3 8 ln |4u+ 1|+ C 4u+ 1 = Ce 4u−8x 3 , (C 6= 0) u = − 1 4 ´C�0ž�) nþ§§�Ï) 4x+ 8y + 1 = Ce 8y−4x 3 , (2x+ 4y − 1 6= 0) (4) w,y = 0´A) if y 6= 0§-u = 1 y2 §K u ′ = −2 y ′ y3 = −2x3 + 2xu = u ′ − 2xu = −2x3 ù´˜‡˜�‚5§§@^úª)� u = x2 + 1 + Cex 2 Ïd y2 = u−1 = (x2 + 1 + Cex 2 )−1 and y = 0 2.)µ (1) -u = x− y§Ku′ = 1− y′ = 1− cosu du+ (cosu− 1)dx = 0 cosu = 1§=u = 2kpi, k ∈ Z´A) if cosu 6= 1 du cosu− 1 + dx = 0 cot u 2 + x = C 22 ∴§�) cot x− y 2 + x = C and x− y = 2kpi, k ∈ Z (2) -u = wv§K§Œz v3(3w + 1)dw + (4w + 2)wv2dv = 0 if v 6= 0, w(4w + 2) 6= 0, 3w + 1 w(4w + 2) dw + dv v = 0 )� w2(2w + 1)v4 = C, (C 6= 0) w,v = 0, w(4w + 2) = 0´þªC�0ž�A)§ Ïd�§�Ï) (wv)2(2wv2 + v2) = C u2(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 = 0 Cþ©l{§)� p = 0 and (h− 1)2 = Cp(h− 2)3 and h = 2 p = 0§=x2 + 2 = 0¿Ø´�§�) ∴ÅÚ£“§��§�) (y2 − x2 − 1)2 = C(y2 − 2x2 − 3)3 and y2 = 2x2 + 3, (y 6= 0) 23 (4) -u = y2, v = x2§K�§z du dv = 2v + 3u− 7 3v + 2u− 8 -w = u− 1, p = v − 2§K�§UYz dw dp = 3w + 2p 2w + 3p -w = ph§K�§UYz p2(2h+ 3)dh+ 2p(h2 − 1)dp = 0 Cþ©l{§)� h+ 1 = Cp4(h− 1)5 and h = 1 and p = 0 p = 0§=x2 − 2 = 0¿Ø´�§�) ∴ÅÚ£“§��§�) y2 + x2 − 3 = C(y2 − x2 + 1)5 and y2 − x2 + 1 = 0 3.)µ (1) -u = xy§K u ′ = xy ′ + y = x(−y2 − 1 4x2 ) + y = 4u− 4u2 − 1 4x , (x 6= 0) if 4u− 4u2 − 1 6= 0§K§z du 4u− 4u2 − 1 = dx 4x )� x2e 4 1−2u = C, (C > 0) 4u− 4u2 − 1 = 0§=u = 1 2 ´§A) ∴§�Ï) x2e 4 1−2xy = C, (C > 0) and xy = 1 2 (2) xy = −1´§x2y′ = x2y2 + xy + 1�˜‡A) 24 -u = y + 1 x §Ku ′ = y ′ − 1 x2 §§Œz u ′ = u2 − u x -z = u−1§K§UYz z ′ − z x + 1 = 0 )� z = Cx− x ln |x| ∴§�Ï) y = −1 x + 1 Cx− x ln |x| and y = − 1 x 4.)µ y = 0´§�˜‡A) if y 6= 0§-u = y−1y′§K u ′ = y−1y ′′ − y−2(y′)2 = y−1 ( −p(x)y′ − q(x)y ) − y−2(y′)2 = −p(x)u− q(x)− u2 =zpk0§ u ′ = −u2 − p(x)u− q(x) 5.)µ�y = y(x)´¤¦­‚§é­‚þ?˜:(x, y)§Ùƒ‚•(1, y ′ )§ •»•(x, y) ∵ƒ‚†•»Y�pi 4 , ∴k§ ∣∣∣∣∣ y ′ − y x 1 + y′ y x ∣∣∣∣∣ = tan(pi4 ) = 1 � y ′ = x+ y x− y or y ′ = y − x x+ y 25 e¡¦)1˜‡§§-y = ux§K§Œz du dx x = 1 + u2 1− u � arctanu− 1 2 ln(1 + u2)− ln |x| = C = arctan y x − 1 2 ln(x2 + y2) = C éu1�‡§§-z = −y§Kz′ = x+z x−z§=1˜‡§ � arctan(−y x )− 1 2 ln(x2 + y2) = C = arctan y x + 1 2 ln(x2 + y2) = C Ïd§¤¦­‚§ arctan y x ± 1 2 ln(x2 + y2) = C 6.)µ �:1  u�:§²‡1º‡��†x¶²1§^=­‚y = y(x)§ dAÛ'X§´wÑL­‚þ?˜:§•»¤3†‚�–��´T? ƒ‚–���2� Kk§ 2y ′ 1− (y′)2 = y x -u = y2§K§z u+ x2 = ( u ′ 2 + x )2 2-v = u+ x2§w,v > 0§K§UYz 4v = (v ′ )2 v ′ = ±2v 12 26 � v = (x+ C)2 = x2 + y2 = (x+ C)2 ¤±´‡�Ԃ y2 = 2Cx+ C2 SK 2-5 1.)µ (1) ∵ ∂P ∂y = 3x2 + 2x+ 3y2, ∂Q ∂x = 2x ∴ 1 Q ( ∂P ∂y − ∂Q ∂x ) = 3 È©Ïfµ(x) = e ∫ 3dx = e3x§¦u§üý§k e3x(3x2y + 2xy + y3)dx+ e3x(x2 + y2)dy = 0 = d ( e3x(x2y + 1 3 y3) ) = 0 e3x(x2y + 1 3 y3) = C (2) ∵ ∂P ∂y = 1, ∂Q ∂x = 2y ∴ 1 P ( ∂Q ∂x − ∂P ∂y ) = 2− 1 y È©Ïfµ(y) = e ∫ (2− 1y )dy = 1 y e2y, (y 6= 0)§¦u§üý§k e2ydx+ e2y ( 2x− 1 y e−2y ) dy = 0 27 = d ( xe2y − ln |y|) = 0 xe2y − ln |y| = C, (y 6= 0) y = 0´A)" (3) §üý¦±xy§Œz (3x2y + 6x)dx+ (x3 + 3y2)dy = 0 †�È©§k x3y + 3x2 + y3 = C (4) (ydx− xdy)− (x2 + y2)dy = 0 1˜|kÈ©Ïfx−2, y−2, (x2 + y2)−1§ 1�|kÈ©Ïf(x2 + y2)−1 ∴�µ = (x2 + y2)−1¦§üý§k y x2 + y2 dx− x 2 + y2 + x x2 + y2 dy = 0 � arctan y x + y = C and y = 0 (5) ∵ ∂P ∂y = 6xy2, ∂Q ∂x = 2xy2 ∴ 1 P ( ∂Q ∂x − ∂P ∂y ) = −4xy2 2xy3 = −2 y È©Ïfµ(y) = e ∫ (− 2y )dy = y−2, (y 6= 0)¦u§üý§k 2xydx+ (x2 − y−2)dy = 0 � x2y + 1 y = C and y = 0 (6) (ydx− xdy) + xy2dx = 0 28 w,µ = y−2´§�˜‡È©Ïf§§Œz 1 y dx− x y2 dy + xdx = 0, (y 6= 0) = x y + x2 2 = C, (y 6= 0) y = 0÷v�§§�´A)" (7) §Œ©u (y3dx− 2xy2dy) + 2x2dy = 0 1˜|kÈ©Ïfy−5ÚÏÈ© x y2 = C 1�|kÈ©Ïfx−2ÚÏÈ©2y = C 8IµÏ錇¼êg1, g2, s.t.÷v y−5g1( x y2 ) = x−2g2(2y) w,g1(t) = t −2, g2(t) = 2t−1÷vþª ∴�§kÈ©Ïfx−2y−1,K§Œz x−2y2dx+ 2(y−1 − x−1y)dy = 0, (xy 6= 0) � −y 2 x + 2 ln |y| = C, (xy 6= 0) x = 0, y = 0´�§�A)§Ïd ln y2 − y 2 x = C and x = 0 and y = 0 (8) ∵ ∂P ∂y = 0, ∂Q ∂x = ex cot y ∴ 1 P ( ∂Q ∂x − ∂P ∂y ) = cot y È©Ïfµ(y) = e ∫ cot ydy = sin y§¦u§üý§k ex sin ydx+ (ex cos y + 2y sin y cos y) dy = 0 29 � ex sin y + 1 4 (sin(2y)− 2y cos(2y)) = C 2.yµ P (x, y)dx+Q(x, y)dy = 0 (2.55) ⇒ e§(2.55)k/Xµ = µ (ϕ(x, y))�È©Ïf§K ∂(µP ) ∂y = ∂(µQ) ∂x = P ∂µ ∂y −Q∂µ ∂x = µ ( ∂Q ∂x − ∂P ∂y ) µ = µ (ϕ(x, y)) ∴ Pµ′ (ϕ(x, y)) ∂ϕ ∂y −Qµ′ (ϕ(x, y)) ∂ϕ ∂x = µ ( ∂Q ∂x − ∂P ∂y ) -f (ϕ(x, y)) = µ ′ (ϕ(x,y)) µ(ϕ(x,y)) §K f (ϕ(x, y)) = ∂Q ∂x − ∂P ∂y P ∂ϕ ∂y −Q∂ϕ ∂x ⇐ e§(2.55)÷v ∂Q ∂x − ∂P ∂y P ∂ϕ ∂y −Q∂ϕ ∂x = f (ϕ(x, y)) -µ(t) = e ∫ f(t)dt§K µ ′ (t) µ(t) = f(t) l µ ′ (ϕ(x, y)) µ (ϕ(x, y)) = ∂Q ∂x − ∂P ∂y P ∂ϕ ∂y −Q∂ϕ ∂x 30 �nŒ� ∂(µP ) ∂y = ∂(µQ) ∂x ∴ µ (ϕ(x, y)) = e ∫ f(ϕ(x,y))dϕ ´§(2.55)�˜‡È©Ïf" (1) òµ = µ(x± y)“\¿‡^‡ªf§�nŒ� ∂P ∂y − ∂Q ∂x Q∓ P = f(x± y) (2) òµ = µ(x2 + y2)“\¿‡^‡ªf§�nŒ� ∂P ∂y − ∂Q ∂x xQ− yP = f(x 2 + y2) (3) òµ = µ(xy)“\¿‡^‡ªf§�nŒ� ∂P ∂y − ∂Q ∂x yQ− xP = f(xy) (4) òµ = µ( y x )“\¿‡^‡ªf§�nŒ� ∂P ∂y − ∂Q ∂x y x2 Q− 1 x P = f( y x ) (5) òµ = µ(xαyβ)“\¿‡^‡ªf§�nŒ� ∂P ∂y − ∂Q ∂x αx−1Q− βy−1P = f(x αyβ) 3.yµ Iy ∂(µP ) ∂y = ∂(µQ) ∂x =y P ∂µ ∂y −Q∂µ ∂x = µ ( ∂Q ∂x − ∂P ∂y ) (3.1) µ = 1 xP+yQ § ∴  ∂µ ∂x = −P+x ∂P∂x +y ∂Q∂x (xP+yQ)2 ∂µ ∂y = −Q+x ∂P ∂y +y ∂Q ∂y (xP+yQ)2 31 “\(3.1)ª§z{§� yQ ∂P ∂y + xQ ∂P ∂x = yP ∂Q ∂y + xP ∂Q ∂x (3.2) �Iy(3.2)ª ¯¢þ§du§´Ùg§§¤±P = ϕ( y x )Q ∴ { ∂P ∂x = − y x2 ϕ ′ ( y x )Q+ ϕ( y x )∂Q ∂x ∂P ∂y = 1 x ϕ ′ ( y x )Q+ ϕ( y x )∂Q ∂y “\(3.2)ª†à left of (3.2) = yQ2 x ϕ ′ + yϕQ ∂Q ∂y + ( −xyQ 2 x2 ϕ ′ + xϕQ ∂Q ∂x ) = yϕQ ∂Q ∂y + xϕQ ∂Q ∂x = yP ∂Q ∂y + xP ∂Q ∂x = right of (3.2) Ïd�yœ 4.yµ (½n6�y²) ∵ µP (x, y)dx+ µQ(x, y)dy = dΦ(x, y) ∴ µg(Φ)P (x, y)dx+ µg(Φ)Q(x, y)dy = g(Φ)dΦ(x, y) = d ∫ g(Φ)dΦ ∴ µ(x, y)g(Φ(x, y))´§�˜‡È©Ïf" (½n2.6�_½n) ∵ µ1, µÑ´‡©§(2.55)�È©Ïf ∴ µ1(Pdx+Qdy) = dψ µ(Pdx+Qdy) = dφ ∴ D[ψ, φ] D[x, y] = ∣∣∣∣∣ ∂ψ∂x ∂φ∂x∂ψ ∂y ∂φ ∂y ∣∣∣∣∣ = ∣∣∣∣∣ µ1P µPµ1Q µQ ∣∣∣∣∣ ≡ 0 32 Ïdψ†φ¼êƒ',l 3¼êf(·)§¦�÷v ψ = f(φ) µ1 µ = dψ dφ = f ′ (φ) , g(φ) ∴ µ1 = µg(φ) �yœ 5.yµ ∵ µ1, µ2Ñ´‡©§(2.55)�È©Ïf ∴ µ1(Pdx+Qdy) = dψ µ2(Pdx+Qdy) = dφ |^þK(J§µ1 = µ2g(φ)§Ù¥g(·)´Œ‡š"¼ê ∴ µ1 µ2 = g(φ) l g ′ (φ)µ2(Pdx+Qdy) = g ′ (φ)dφ = dg(φ) q∵ µ1 µ2 Øð~ê§∴ g′(φ)Øð" Ïdg ′ (φ)µ2´§�˜‡È©Ïf ƒA�ÏÈ©g(φ) = µ1 µ2 = C§�yœ SK 2-6 1.)µ (1) ∵ x2 + y2 = Cx ∴ 2x+ 2yy′ = C = x 2 + y2 x ∴§���;‚x§÷v 2x− 2y 1 y′ = x2 + y2 x 33 = 2xydx+ (y2 − x2)dy = 0 kÈ©Ïfµ = y−2,¦§üý§� 2x y dx+ ( 1− x 2 y2 ) dy = d ( x2 y + y ) = 0 =k x2 + y2 = Ky (2) ∵ xy = C ∴ y + xy′ = 0 ∴§���;‚x§÷v y − x y′ = 0 = xdx− ydy = 0 =k x2 − y2 = K (3) ∵ y2 = ax3 ∴ 2yy′ = 3ax2 = 3y 2 x ∴§���;‚x§÷v −2y y′ = 3y2 x = 2xdx+ 3ydy = 0 =k x2 + 3 2 y2 = K (4) ∵ x2 + C2y2 = 1 34 ∴ 2x+ 2C2yy′ = 2x+ 21− x 2 y y ′ = 0 ∴§���;‚x§÷v x− 1− x 2 y 1 y′ = 0 = 1− x2 x dx− ydy = 0 =k x2 + y2 − lnx2 = K 2.)µ (1) ∵ x− 2y = C ∴ H(x, y) = y′ = 1 2 ∴¤¦­‚x§÷v dy dx = H + tan pi 4 1−H tan pi 4 = 3 =k y = 3x+K (2) ∵ xy = C ∴ H(x, y) = y′ = −y x ∴¤¦­‚x§÷v dy dx = H + tan pi 4 1−H tan pi 4 = x− y x+ y = (y − x)dx+ (x+ y)dy = 0 =k x2 − y2 − 2xy = K 35 (3) ∵ y = x ln ax ∴ H(x, y) = y′ = ln ax+ 1 = 1 + y x ∴¤¦­‚x§÷v dy dx = H + tan pi 4 1−H tan pi 4 = −2x+ y y -y = ux§K§Œz u+ 1 + 2 u + x du dx = 0 � ln ( (u+ 1 2 )2 + 7 4 ) − 2√ 7 arctan 2u+ 1√ 7 + lnx2 = K = ln(y2 + xy + 2x2)− 2√ 7 arctan 2y + x√ 7x = K (4) ∵ y2 = 4ax ∴ H(x, y) = y′ = 4a 2y = y2 2xy = y 2x ∴¤¦­‚x§÷v dy dx = H + tan pi 4 1−H tan pi 4 = 2x+ y 2