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201 Bài tập Phương trình vi phân

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  1. www.VNMATH.com 1 . . BAI` TAˆ. P PHUONG TR`INH VI PHANˆ . . 1) Gia'i phuong trnh: 2xy0y” = y02 − 1 HD gia’i: D- a. t y0 = p : 2xpp0 = p2 − 1 . 2pdp dx √ Voi x(p2 − 1) 6= 0 ta co: = ⇔ p2 − 1 = C ⇔ p = ± C x + 1 p2 − 1 x 1 1 dy √ 2 3 p = = C1 + 1 ⇒ y = (C1x + 1) 2 + C2 dx 3C1 . . √ 2) Gia'i phuong trnh: y.y” = y0 dp . . . √ dp HD gia’i: D- a. t y0 = p ⇒ y” = p (hamtheo y). Phuong trnhtro' thanh: yp = p dy dy . . . . . dy √ dy √ Voi p 6= 0 ta duo. c phuong trnh: dp = √ ⇒ p = 2 y + C ⇔ = 2 y + C ⇒ y 1 dx 1 dy dx = √ 2 y + C1 . √ C1 √ Tu donghi^e.m t^o'ng quat: x = y − ln |2 y + C | + C 2 1 2 Ngoaira y = c: hang cu~ng langhi^e.m. . . 3) Gia'i phuong trnh: a(xy0 + 2y) = xyy0 HD gia’i: a(xy0 + 2y) = xyy0 ⇒ x(a − y)y0 = −2ay . . . . . . . a − y 2a N^eu y 6= 0, ta cophuong trnhtuong duong voi dy = − dx ⇔ x2ayae−y = C y x Ngoaira y = 0 cu~ng langhi^e.m. . . 4) Gia'i phuong trnh: y” = y0ey dp . . dp HD gia’i: D- a. t y0 = p ⇒ y” = p thay vaophuong trnh: p = pey dy dy dp dy dy Vo.i y y y p 6= 0 : = e ⇔ p = e + C1 ⇒ = e + C1 ⇔ y = dx dy dx e + C1 y y y Vo.i ta co: R dy 1 R e + C1 − e 1 R e dy y C1 6= 0 y = y dy = (y − y ) = − e + C1 C1 e + 1 C1 e + C1 C1 1 y ln(e + C1) C1  −y −e nˆe´u C1 = 0 nhu. v^ay: R dx  . y = 1 y e + C1  (y − ln |e + C1|) nˆe´u C1 6= 0. C1 Ngoaira y = C : hang lam^o. t nghi^e.m . . . 5) Gia'i phuong trnh: xy0 = y(1 + ln y − ln x) voi y(1) = e
  2. 2 www.VNMATH.com . . . y y . . HD gia’i: D- ua phuong trnh v^e: y0 = (1 + ln ), da. t y = zx duo. c: xz0 = z ln z x x dz dx y • z ln z 6= 0 ⇒ = ⇒ ln z = Cx hay ln = Cx ⇔ y = xeCx z ln z x x y(1) = e → C = 1. V^a. y y = xex . . 6) Gia'i phuong trnh: y”(1 + y) = y02 + y0 dz . . dz dy HD gia’i: D- a. t y0 = z(y) ⇒ z0 = z thay vaophuong trnh: = dy z + 1 y + 1 dy ⇒ z + 1 = C1(y + 1) ⇒ z = C1y + C1 − 1 ⇔ = dx (∗) C1y + C1 − 1 • C1 = 0 ⇒ (∗) cho y = C − x 1 • C1 6= 0 ⇒ (∗) cho ln |C1y + C1 − 1| = x + C2 C1 Ngoaira y = C langhi^e.m. ' 1 Tom la.i nghi^e.m t^ong quat: y = C, y = C − x; ln |C1y + C1 − 1| = x + C2 C1 . . 2 7) Gia'i phuong trnh: y0 = y2 − x2 HD gia’i: Bi^en d^o'i (3) v^e da.ng: x2y0 = (xy)2 − 2 (∗) D- a. t z = xy ⇒ z0 = y + xy0 thay vao (∗) suy ra: dz dx rz − 1 xz0 = z2 + z − 2 ⇔ = ⇔ 3 = Cx z2 + z − 2 x z + x xy − 1 V^a. y TPTQ: = Cx3. xy + 2 . . 8) Gia'i phuong trnh: yy” + y02 = 1 dz HD gia’i: D- a. t y0 = z(y) ⇒ y” = z. dy . . z dy C1 Bi^en d^o'i phuong trnh v^e: dz = ⇔ z2 = 1 + 1 − z2 y y2 dy r C dy ⇒ = ± 1 + 1 ⇔ ± R = dx ⇒ y2 + C = (x + C )2 dx y2 r C 1 2 1 + 1 y2 ' 2 2 Nghi^e.m t^ong quat: y + C1 = (x + C2) . . √ 9) Gia'i phuong trnh: 2x(1 + x)y0 − (3x + 4)y + 2x 1 + x = 0 3x + 4 1 HD gia’i: y0 − .y = −√ ; x 6= 0, x 6= −1 2x(x + 1) x + 1 . . Nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at: dy 3x + 4 2 1 Cx2 R = R dx = R ( − )dx ⇔ y = √ y 2x(x + 1) x 2(x + 1) x + 1
  3. www.VNMATH.com 3 1 1 Bi^en thi^enhang s^o: C0 = − ⇒ C = − + ε. x2 x x2 1 V^a. y nghi^e.m t^o'ng quat: y = √ ( + ε) x + 1 x ( . . y(0) = 0 10) Gia'i phuong trnh: y” = e2y thoa' y0(0) = 0 dz . . . dz z2 e2y HD gia’i: D- a. t z = y0 → y” = z. phuong trnh tro' thanh z. = e2y ⇔ = + ε dy dy 2 2 1 . y0(0) = y(0) = 0 ⇒ ε = − . V^a. y z2 = e2y − 1. Tu do: 2 dy √ Z dy √ z = = e2y − 1 ⇒ √ = x + ε. d¯ˆo’i biˆe´n t = e2y − 1 dx e2y − 1 √ arctg e2y − 1 = x + ε 1 y(0) = 0 ⇒ ε = 0. V^a. y nghi^e.m ri^engthoa' di^eu ki^e.n d^e bai: y = ln(tg2x + 1). 2 . . 11) Tmnghi^e.m ri^engcu'a phuong trnh: xy0 + 2y = xyy0 thoa' ma~n di^eu ki^e.n d^au y(−1) = 1. . . . HD gia’i: Vi^et phuong trnh la.i: x(1 − y)y0 = −2y; do y(−1) = 1 n^en y 6≡ 0.D- ua v^e . . 1 − y dx phuong trnh tach bi^en: dy = −2 y x . . 1 tch ph^ant^o'ng quat: x2ye−y = C. Thay di^eu ki^e.n vaota duo. c C = . V^a. y tch ph^an e ri^engc^an tmla: x2ye1−y = 1. . . 12) Bang cach da.t y = ux, ha~y gia'i phuong trnh: xdy − ydx − px2 − y2dx = 0. (x > 0) . . . . HD gia’i: D- a. t y = ux; du = udx + xdu thay vaophuong trnhvagia'n uoc x: xdu − √ . . . 1 − u2dx = 0. Ro~ rang u − ±1 langhi^e.m. khi u 6≡ ±1 dua phuong trnhv^e tach bi^en: du dx = . TPTQ: arcsin u − ln x = C (do x > 0). 1 − u2 x . . y V^a. y NTQ cu'a phuong trnh: y = ±x; arcsin = ln x + C. x . . 13) Tmnghi^e.m ri^engcu'a phuong trnh: xy0 = px2 − y2 + y thoa' ma~n di^eu ki^e.n d^au y(1) = 0. HD gia’i: r p y2 y xy0 = x2 − y2 + y ⇐⇒ y0 = 1 − + x2 x y da. t u = hay y = ux suy ra y0 = xu0 + u x . . √ du dx phuong trnh thanh: xu0 = 1 − u2 ⇐⇒ √ = 1 − u2 x
  4. 4 www.VNMATH.com ⇐⇒ arcsin u = ln Cx thoa' ma~n di^eu ki^e.n d^au y(1) = 0 khi C = 1. V^a. y nghi^e.m y = ±x. . . 14) Tmnghi^e.m ri^engcu'a phuong trnh: y0 sin x = y ln y π thoa' ma~n di^eu ki^e.n d^au y( ) = e. 2 HD gia’i: dy dx y0 sin x = y ln y ⇐⇒ = y ln y sin x x x C tan ⇐⇒ ln y = C tan ⇐⇒ y = e 2 2 x π tan thoa' ma~n di^eu ki^e.n d^au y( ) = e khi C = 1. V^a. y y = e 2 . 2 . . 15) Tmnghi^e.m ri^engcu'a phuong trnh: (x + y + 1)dx + (2x + 2y − 1)dy = 0 thoa' ma~n di^eu ki^e.n d^au y(0) = 1. HD gia’i: D- a. t x + y = z =⇒ dy = dz − dx . . phuong trnhthanh: (2 − z)dx + (2z − 1)dz = 0; gia'i ra x − 2z − 3 ln |z − 2| = C. V^a. y x + 2y + 3 ln |x + y − 2| = C thoa' ma~n di^eu ki^e.n d^au y(0) = 1 khi C = 2. 1 16) Bang cach da.t y = r^oi da.t z = ux,ha~y gia'i . . z phuong trnh: (x2y2 − 1)dy + 2xy3dx = 0 1 . . . . HD gia’i: D- a. t y = duo. c: (z2 − x2)dz + 2zxdx = 0; r^oi da. t z = ux, duo. c z (u2 − 1)(udx + xdu) + 2udx = 0 dx u2 − 1 ⇐⇒ + du = 0 x u3 + u u2 + 1 x(u2 + 1) ⇐⇒ ln |x| + ln = ln C ⇐⇒ = C |u| u 1 . . thay u = duo. c nghi^e.m 1 + x2y2 = Cy. xy . . 17) Tmnghi^e.m t^o'ng quat cu'a phuong trnhsau: y0 − xy = x + x3 HD gia’i: . . D- ^aylaphuong trnh tuy^en tnh c^ap 1 vaconghi^e.m t^o'ng quat la 2 x2 x y = Ce 2 . + 1 2 .
  5. www.VNMATH.com 5 . . 18) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 − y = y2. . . HD gia’i: D- ^aylaphuong trnhtach bi^en vaconghi^e.m t^o'ng quat la y ln | | = x + C. y + 1 . . y 19) Tmnghi^e.m cu'a cac phuong trnhsau: y0 + = ex x HD gia’i: . . C ex D- ^aylaphuong trnh tuy^en tnh c^ap 1 vaconghi^e.m t^o'ng quat la y = + ex − . x x . . 20) Tmnghi^e.m cu'a cac phuong trnhsau: y0 − y = y3. . . HD gia’i: D- ^aylaphuong trnhtach bi^en vaconghi^e.m t^o'ng quat la C + x = ln |y| − arctgy. . . y y . π 21) Gia'i phuong trnh: y0 = + sin , voi y(1) = x x 2 . . . HD gia’i: y = zx ⇒ y0 = z0x + z, phuong trnh tro' thanh: dz dx z z z0x = sin x ⇔ = ⇔ ln |tg | = ln |x| + ln C ⇔ tg = Cx sin z x 2 2 y π V^a. y nghi^e.m t^o'ng quat: tg = Cx; y(1) = ⇒ C = 1. 2x 2 y V^a. y: tg = x. 2x . . y y 22) Gia'i phuong trnh: (x − y cos )dx + x cos dy = 0 x x y . . . . . HD gia’i: D- a. t = z ⇒ y0 = z0x + z phuong trnh duo. c dua v^e da.ng: x Z dx x cos z.z0 + 1 = 0 ⇔ cos zdz = − + C ⇔ sin z = − ln |x| + C x y V^a. y TPTQ: sin = − ln |x| + C x . . 23) Gia'i phuong trnh: (y02 − 1)x2y2 + y0(x4 − y4) = 0 . . . . HD gia’i: Laphuong trnh da' ng c^ap nhung gia'i khaphuc ta.p.
  6. 6 www.VNMATH.com . . . y2 x2 Xem phuong trnh b^a. c hai d^oi voi y0: 4 = (x4 + y4)2 ⇒ y0 = ; y0 = − . 1 x2 2 y2 . ' x 3 3 Tu docohai ho. nghi^e.m t^ong quat: y = ; x + y = C2 C1x + 1 . . 24) Gia'i phuong trnh: y2 + x2y0 = xyy0 y2 Vi^et phu.o.ng trnhlai 0 x2 d^aylaphu.o.ng trnhthu^an nh^at, gia'i HD gia’i: . y = y x − 1 . . ' 2 y ra duo. c nghi^e.m t^ong quat: y = Cxe x . . 25) Tmnghi^e.m ri^engcu'a phuong trnh: (x + y − 2)dx + (x − y + 4)dy = 0 thoa' ma~n di^eu ki^e.n d^au y(1) = 0. ( x = u − 1 . . . . HD gia’i: D- a. t thay vaophuong trnh duo. c: y = v + 3. . . (u + v)du + (u − v)dv = 0, d^aylaphuong trnhthu^an nh^at cotch ph^ant^o'ng quat la: u2 + 2uv − v2 = C. . . V^a. y tch ph^ant^o'ng quat cu'a phuong trnhban d^au la: x2 + 2xy − y2 − 4x + 8y = C . . 26) Gia'i phuong trnh (x + y − 2)dx + (x − y + 4)dy = 0. ( x = X − 1 . . HD gia’i: D- a. t , phuong trnh thanh: y = Y + 3 (X + Y )dX + (X − Y )dY = 0 . . . dX 1 − u da. t Y = uX dua phuong trnh v^e + du = 0. X 1 + 2u − u2 Gia'i ra X2(1 + 2u − u2) = C hay x2 + 2xy − y2 − 4x + 8y = C. . . 2xy 27) Tmtch ph^ant^o'ng quat cu'a phuong trnhsau: b) y0 = . x2 − y2 . . y . . HD gia’i: D- ^aylaphuong trnhda' ng c^ap, ta da. t z = . Khi dophuong trnhtr^en z . z(1 + z2) 1 2z dx . . tro' thanh xz0 = . Hay ( − )dz = . Suy ra nghi^e.m cu'a phuong trnh 1 − z2 z 1 + z2 x z nayla = Cx, C 6= 0. 1 + z2 ' . . ~ 2 2 V^a. y nghi^e.m cua phuong trnh da cho la x + y = C1y, C1 6= 0. . . 2x + y − 1 28) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 = . 4x + 2y + 5 . . . HD gia’i: D- a. t u = 2x + y phuong trnh dua v^e da.ng du 5u + 9 = . dx 2u + 5
  7. www.VNMATH.com 7 . . . . Gia'i phuong trnhnayta duo. c nghi^e.m 10u + 7 ln |5u + 9| = 25x + C. . . V^a. y nghi^e.m cu'a phuong trnh da~ cho la 10y + 7 ln |10x + 5y = 9| − 5x = C. . . 29) Tmtch ph^ant^o'ng quat cu'a cac phuong trnhsau: (x − y + 4)dy + (y + x − 2)dx = 0 . . . . . HD gia’i: D- ^aylaphuong trnhdua v^e da.ng da' ng c^ap duo. c bang cach da. t x = . . dv u + v . . . . u + 1, y = v − 3, ta duo. c = . Gia'i phuong trnhta conghi^e.m cu'a phuong du −u + v trnhla v2 − 2uv − v2 = C. ' . . ~ 2 2 V^a. y nghi^e.m cua phuong trnh da cho la y − x − 2xy − 8y + 4x = C1. a) Tmmi^en matrong donghi^em cu'a baitoan Cauchy cu'a phu.o.ng trnh 30) . √ sau d^ayt^on ta.i vaduy nh^at y0 = x − y. . . b) Tmtch ph^ant^o'ng quat cu'a cac phuong trnhsau: (x2 − y2)dy − 2xydx = 0. HD gia’i: a) Baitoan Cauchy coduy nh^at nghi^e.m trong mi^en . D = {(x, y) ∈ R2|x − y ≥ δ} voi δ > 0 tuyy. . . . dy xy . . b) D- ua phuong trnhv^e da.ng = .D- ^aylaphuong trnhda' ng c^ap, ta da. t dx x2 − y2 y . . . z = . Khi dophuong trnh tr^entro' thanh x z(1 + z2) xz0 = . 1 − z2 1 2z dx Hay ( − )dz = . z 1 + z2 x . . z Suy ra nghi^e.m cu'a phuong trnh nayla = Cx, C 6= 0. 1 + z2 ' . . ~ 2 2 V^a. y nghi^e.m cua phuong trnh da cho la x + y = C1y, C1 6= 0. . . 31) a) Chung minh rang h^e. cac vecto {e2x, xe2x, x2} lah^e. d^o.c l^a.p tuy^en tnh. . . b) Tmtch ph^ant^o'ng quat cu'a phuong trnhsau: (x − y)dy − (x + y)dx = 0; HD gia’i: a) Dungdi.nh ngh~a ki^e'm tra h^e. d^o. c l^a. p tuy^en tnh . . . . x + y . . b) D- ua phuong trnh v^e da.ng y0 = .D- ^aylaphuong trnhda' ng c^ap, ta da. t x − y y . . . z = . Khi dophuong trnh tr^entro' thanh x 1 + z2 xz0 = . 1 − z . . . . Gia'i phuong trnhnayta duo. c p arctg y x2 + y2 = Ce x . . . 32) a) Chung minh rang h^e. cac vecto {cos2 2x, sin2 2x, 2} lah^e. phu. thu^o.c tuy^en tnh. . Tnh di.nh thuc Wronski cu'a chung. . . b) Tmtch ph^ant^o'ng quat cu'a phuong trnhsau: (x − 2y + 1)dy − (x + y)dx = 0.
  8. 8 www.VNMATH.com HD gia’i: a) H^e. nayphu. thu^o. c tuy^en tnh v 2 cos2 2x + 2 sin2 2x − 2 = 0. . . . . . b) Phuong trnhnaycoth^e' dua v^e da.ng da' ng c^ap, ta duo. c x + y y0 = . x − 2y + 1 1 1 . . . D- a. t u = x − , v = y + , khi dophuong trnh tr^entro' thanh 3 3 u + v v0 = . u − 2v √ . . . . √ √1 arctg( 2 u ) Gia'i phuong trnhnayta duoc 2 2 2 v . u + 2√v = Ce . √1 3x−1 p 2 2 arctg( 2 3y+1 ) Hay (3x − 1) + 2(3y + 1) = C1e 2 . . . 33) Gia'i phuong trnh: y2 + x2y0 = xyy0 . . HD gia’i: Phuong trnhthu^an nh^at: da. t y = zx → y0 = z0x + z . . . z − 1 dx Phuong trnh tro' thanh dz = → z − ln |z| = ln |x| + C z x y y − ln | | = ln |x| + C x x . . 34) Gia'i phuong trnh y2 + x2y0 = xyy0. y2 Vi^et phu.o.ng trnhlai 0 x2 d^aylaphu.o.ng trnhthu^an nh^at, gia'i HD gia’i: . y = y x − 1 . . ' 2 y ra duo. c nghi^e.m t^ong quat: y = Cxe x . . 35) Gia'i phuong trnh: y” cos y + (y0)2 sin y = y0 HD gia’i: y = C : hang lam^o. t nghi^e.m. dp y 6= C (hang). D- a. t y0 = p ⇒ y” = p (hamtheo y) dy dp . . thay vao(2): cos y + p sin y = 1: phuong trnhtuy^en tnh. dy . . Phuong trnh thu^an nh^at conghi^e.m t^o'ng quat: p = C cos y.    . . bi^en thi^enhang s^o duo. c C = tgy + C1. . dy dy tu do p = = sin y + C1 cos y ⇔ = dx dx sin y + C1 cos y y r 1 1 tg + 1 + − 2  1 2 C1 C1 tch ph^andid^en: ln = x + C2 p 2 r C1 + 1 y 1 1 −tg + 1 + 2 + 2 C1 C1 . . 1 36) Gia'i phuong trnh: y0 + = 0 2x − y2 1 . . HD gia’i: Coi x = x(y) lahamcu'a y ta co: y0 = thay vaophuong trnh: x0
  9. www.VNMATH.com 9 1 1 . . + = 0 ⇔ x0 + 2x = y2 : phuong trnh tuy^en tnh. x0 2x − y2 . . Nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at: x = Ce−2y 1 1 1 Bi^en thi^enhang s^o: C0(y) = y2e2y ⇒ C(y) = y2e2y − ye2y + e2y + C 2 2 4 . . 1 1 1 V^a. y nghi^e.m t^o'ng quat cu'a phuong trnh: x = Ce−2y + y2 − y + 2 2 4 . . 37) Gia'i phuong trnh: xy” = y0 + x2 . HD gia’i: D- a. t y0 = p, (1) tro' thanh: xp0 − p = x2 tuy^en tnh . . Nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at: p = Cx    Bi^en thi^enhang s^o → C(x) = x + C1 dy x3 x2 Suy ra: = x(x + C ) → y = + C . + C dx 1 3 1 2 2 . . 38) Gia'i phuong trnh: y02 + yy” = yy0 . . . . . . . dp HD gia’i: D- a. t p = y0(p 6= 0), phuong trnh tuong duong voi: p2 + yp = yp dy dp . . . dp p ⇔ p + y = y, xet y 6= 0 dua phuong trnh v^e: + = 1 (tuy^en tnh) dy dy y . . C NTQ cu'a phuong trnhthu^an nh^at: p = , bi^en thi^enhang s^o y y2 ⇒ C(y) = + C 2 1 2 2 Nhu. v^ay: y + 2C1 dy y + 2C1 2ydy . p = ⇒ = ⇒ 2 = dx 2y dx 2y y + 2C1 2 x ⇒ y = A1e + A2.  0 0 0 0 x x 2 x Chuy: V^e trai (yy ) = yy ⇔ yy = C1e ⇔ ydy = C1e dx ⇔ y = 2C1e + C2 . . . 39) Gia'i phuong trnh: yey = y0(y3 + 2xey) voi y(0) = −1 1 . . 2 HD gia’i: y0 = bi^en d^o'i phuong trnh v^e: x0 − x = y2e−y x x0y y Nghi^e.m t^o'ng quat: x = y2(C − e−y) y(0) = −1 ⇒ C = e. V^a. y x = y2(e − e−y) . . 40) Gia'i phuong trnh: xy” = y0 + x . . . 1 HD gia’i: D- a. t y0 = p; phuong trnh tro' thanh: p0 − p = 1 x '    Nghi^e.m t^ong quat: p = Cx bi^en thi^enhang s^o: C = ln |x| + C1
  10. 10 www.VNMATH.com dy Z ⇒ p = = (ln |x| + C )x ⇒ y = (ln |x| + C )xdx + C dx 1 1 2 x2 x2 = C x2 + ln |x| − + C 1 2 4 2 . . 41) Gia'i phuong trnh: y0 + xy = x3 2 ' . . − x HD gia’i: Nghi^e.n t^ong quat cu'a phuong trnh thu^an nh^at y = Ce 2 2 2 − x bi^en thi^enhang s^o: C(x) = (x − 2)e 2 + ε 2 ' − x 2 V^a. y nghi^e.m t^ong quat: y = εe 2 + x − 2. . . 42) Gia'i phuong trnh: (x2 − y)dx + xdy = 0 . . . . HD gia’i: Phuong trnhvi^et la.i: xy0 −y = −x2, phuong trnhthu^an nh^at: xy0 −y = 0 conghi^e.m t^o'ng quat: y = Cx bi^en thi^enhang s^o suy ra C = −x + ε V^a. y nghi^e.m t^o'ng quat : y = −x2 + εx . . 2 3 . 43) Gia'i phuong trnh: y0 − y = voi y(1) = 1 x x2 . . 3 1 HD gia’i: Phuong trnhtuy^en tnh: y = Cx2; C0 = ⇒ C = − + ε x4 x3 1 y = εx2 − ; y(1) = 1 ⇒ ε = 2 x 1 V^a. y nghi^e.m t^o'ng quat: y = 2x2 − x . . 44) Gia'i phuong trnh: (x + 1)(y0 + y2) = −y . . 1 HD gia’i: Xet y 6= 0, bi^en d^o'i phuong trnhv^e da.ng y0 + .y = −y2 x + 1 1 z0 . . . 1 D- a. t = z ⇒ y0 = − = −y2z0 dua phuong trnh v^e z0 − .z = 1. y z2 x + 1 ' ' . .      Nghi^e.m t^ong quat cua phuong trnhthu^an nh^at: z = C1(x + 1) bi^en thi^enhang s^o C1 = ln |x + 1| + ε. V^a. y nghi^e.m: z = (x + 1)(ln |x + 1| + ε) ngoaira y = 0 cu~ng langhi^e.m. 1 V^a. y nghi^e.m t^o'ng quat: y = va y = 0 nghi^e.m kdi (x + 1)(ln |x + 1| + ε) . . 1 45) Gia'i phuong trnh: 2xy0 + y = 1 − x . . . 1 1 . . HD gia’i: D- ua phuong trnhv^e da.ng y0 + y = phuong trnhtuy^en 2x 2x(1 − x) tnh c^ap 1
  11. www.VNMATH.com 11 C Nghi^e.m t^o'ng quat: y = √ , bi^en thi^enhang s^o: x √ √ x 1 x + 1 C0(x) = ⇒ C = ln |√ | + ε 2x(1 − x) 2 x − 1 √ 1 1 x + 1  V^a. y nghi^e.m t^o'ng quat: y = √ ln |√ | + ε x 2 x − 1 . . 46) Gia'i phuong trnh: xy0 − y = x2 sin x y . . HD gia’i: y0 − = x sin x, phuong trnhtuy^en tnh. NTQ: y = Cx bi^en thi^enhang x s^o: Nghi^e.m t^o'ng quat: y = (C − cos x)x . . 47) Gia'i phuong trnh: y0 cos2 x + y = tgx thoa' y(0) = 0 . . HD gia’i: Phuong trnhtuy^en tnh → NTQ y = Ce−tgx; y = tgx − 1 (m^o. t nghi^e.m ri^eng) ⇒ NTQ: y = Ce−tgx + tgx − 1 y(0) = 0 ⇒ C = 1. V^a. y nghi^e.m ri^engc^an tm: y = tgx − 1 + e−tgx. . . √ 48) Gia'i phuong trnh: y0 1 − x2 + y = arcsin x thoa' y(0) = 0 . . HD gia’i: Nghi^e.m t^o'ng quat cu'a phuong trnhtuy^en tnh thu^an nh^at: y = Ce−arcsinx D^e~ th^ay nghi^e.m ri^eng: y = arcsinx − 1 ⇒ NTQ: y = Ce−arcsinx + arcsinx − 1 y(0) = 0 ⇒ C = 1 ⇒ nghi^e.m ri^engc^an tm: y = e−arcsinx + arcsinx − 1 . . 1 49) Tmnghi^e.m ri^engcu'a phuong trnh: y0 = 2x − y2 thoa' ma~n di^eu ki^e.n d^au y(1) = 0. 1 . . HD gia’i: Xem x la^a'n ham,thay y0 = , phuong trnh thanh x0 1 1 = ⇐⇒ x0 − 2x = −y2 x0 2x − y2 . . . . D- ^aylaphuong trnhtuy^en tnh c^ap m^o. t, nghi^e.m t^o'ng quat cu'a phuong trnhtuy^en . . . . . tnh thu^an nh^at tuong ung la x = Ce−2y. Bi^en thi^enhang s^o duo. c NTQ: y2 y 1 x = Ce−2y + − + 2 2 4 3 thoa' ma~n di^eu ki^e.n d^au y(1) = 0 khi C = . 4 3 y2 y 1 V^a. y nghi^e.m tho'a ma~n di^eu ki^e.n d^au: x = e−2y + − + . 4 2 2 4
  12. 12 www.VNMATH.com . . z . . 50) Gia'i phuong trnhsau d^ay, bi^et rang sau khi da.t y = , ta nh^a.n duo. c x2 . . 1 m^o.t phuong trnhvi ph^anc^ap hai com^o.t nghi^e.m ri^eng y∗ = ex: 2 x2y00 + 4xy0 + (x2 + 2)y = ex. z0x − 2z z00x2 − 4z0x + 6z . . HD gia’i: D- a. t y = zx2 =⇒ y0 = ; y00 = . Phuong trnhthanh x3 x4 ex . . : z00 + z = ex, com^o. t nghi^e.m ri^engla y∗ = , NTQ cu'a phuong trnhthu^an nh^at: 2 ' . .  z = C1 cos x + C2 sin x. V^a. y NTQ cua phuong trnh ban d^au la: cos x sin x ex y = C + C + 1 x2 2 x2 2x2 . . 51) Tmnghi^e.m ri^engcu'a phuong trnh: yey = y0(y3 + 2xey) thoa' ma~n di^eu ki^e.n d^au y(0) = −1. 1 . . 2 HD gia’i: Xem x la^a'n ham,thay y0 = , phuong trnhthanh x0 − x = y2e−y. x0 y . . . . . C NTQ cu'a phuong trnhtuy^en tnh thu^an nh^at tuong ung la x = ; bi^en thi^enhang y . . . C 1 s^o duo. c C(y) = −e−y + C. Nhu v^a. y NTQ la x = − . Thay di^eu ki^e.n d^au xac di.nh y yey . . 1 . duo. c C = . Tu doKL. e . . 52) Tmnghi^e.m cu'a phuong trnh y0 − y = cos x − sin x. tho'a di^eu ki^e.n y bi. cha.n khi x → ∞ . . HD gia’i: Gia'i phuong trnh tuy^en tnh ra y = Cex + sin x tho'a di^eu ki^e.n y bi. cha. n khi x → ∞ khi C = 0 . . 53) Tmnghi^e.m ri^engcu'a phuong trnh: y0 + sin y + x cos y + x = 0 π thoa' ma~n di^eu ki^e.n d^au y(0) = . 2 HD gia’i: y y y y0 + sin y + x cos y + x = 0 ⇐⇒ y0 + 2 sin cos + x.2 cos2 = 0 2 2 2 y0 y ⇐⇒ y + tan + x = 0 2 cos2 2 2 y y0 dat 0 , phu.o.ng trnhthanhphu.o.ng trnhtuy^en tnh . z = tan =⇒ z = y 2 2 cos2 2 z0 + z = −x. Gia'i ra: z = 1 − x + Ce−x π thoa' ma~n di^eu ki^e.n d^au y(0) = khi C = 0. V^a. y nghi^e.m ri^eng y = 2 arctan(1 − x). 2
  13. www.VNMATH.com 13 . . x 54) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 − x tan y = cos y . . . HD gia’i: D- a. t z = sin y, khi dophuong trnhda~ cho tro' thanh z0 − xz = x. D- ^ayla . . ' x2 phuong trnh tuy^en tnh c^ap 1 vaconghi^e.m t^ong quat la z = Ce 2 − 1. . . x2 V^a. y nghi^e.m cu'a phuong trnh da~ cho la sin y = z = Ce 2 − 1 . . 55) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 − xy = x HD gia’i: - . . ' 1 x2 D^aylaphuong trnh tuy^en tnh c^ap 1 vaconghi^e.m t^ong quat la y = Ce 2 − 1. . . y √ 56) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 + = x y. x . . HD gia’i: D- ^aylaphuong trnhBernoulli vaconghi^e.m t^o'ng quat la √ C 1 y = √ + x2. x 5 . . y 57) Tmnghi^e.m cu'a cac phuong trnhsau: y0 − = x3 x HD gia’i: . . D- ^aylaphuong trnh tuy^en tnh c^ap 1 vaconghi^e.m t^o'ng quat la 1 y = Cx + x4. 3 . . 58) Tmnghi^e.m cu'a cac phuong trnhsau: y0 − y = y2. HD gia’i: . . D- ^aylaphuong trnh Bernoulli vaconghi^e.m t^o'ng quat la 1 y2 = . Ce−2x − 1 . . y 59) Tmnghi^e.m cu'a cac phuong trnhsau: y0 + = sin x x HD gia’i: . . D- ^aylaphuong trnh tuy^en tnh c^ap 1 vaconghi^e.m t^o'ng quat la C sin x y = + − cos x. x x
  14. 14 www.VNMATH.com . . √ 60) Tmnghi^e.m cu'a cac phuong trnhsau: y0 − y = x y. HD gia’i: . . D- ^aylaphuong trnh Bernoulli vaconghi^e.m t^o'ng quat la √ 1 x y = Ce 2 − x − 2. . . 2 61) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 + 2xy = xe−x HD gia’i: D- ^aylaphu.o.ng trnh vi ph^antuy^en tnh c^ap 1. 2 x 2 Nghi^e.m t^o'ng quat la y = (C + )e−x . 2 . . y √ 62) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 − 4 = x y. x . . HD gia’i: D- ^aylaphuong trnhBernoulli vaconghi^e.m la √ 1 y = ln x + Cx2. 2 . . 63) a) Tmmi^en matrong donghi^e.m cu'a baitoan Cauchy cu'a phuong trnhsau d^ayt^on ta.i vaduy nh^at y0 = y + 3x.  1 y” − y0 = x b) Tmnghi^e.m cu'a baitoan Cauchy sau d^ay x y(x = 1) = 1 va` y0(x = 1) = 2. HD gia’i: . . a) D- ^aylaphuong trnhtuy^en tnh c^ap 1 tho'a di.nh lydi^eu ki^e.n t^on ta.i duy nh^at nghi^e.m tr^en R2. . . y0 . . b) Gia'i phuong trnh y” − = x, ta duo. c nghi^e.m t^o'ng quat x x2 y = C + C x + . 1 2 2 V^a. y nghi^e.m cu'a baitoan Cauchy la 1 x2 y = − + x + . 2 2 . . 64) Tmnghi^e.m cu'a phuong trnhsau: y0 + ytgx = cos x HD gia’i: D- ^aylaphu.o.ng trnh vi ph^antuy^en tnh c^ap 1. Nghi^e.m t^o'ng quat la: y = (C + x) cos x.
  15. www.VNMATH.com 15 . . y ex 65) Tmnghi^e.m cu'a phuong trnhsau: y0 + = x( )y2. x ex + 1 HD gia’i: . . . . D- ^aylaphuong trnhvi ph^anBernoulli vaconghi^e.m t^o'ng quat cu'a phuong trnhla 1 y = . Cx − x ln(ex + 1) . . 66) Gia'i phuong trnh: (x + 1)y” + x(y0)2 = y0 . . . . . . HD gia’i: D- a. t y0 = p, phuong trnh tro' thanhphuong trnh Bernouili (voi x 6= −1) 1 x p0 − p = − p2 (∗) x + 1 x + 1 . . . D- a. t z = p−1 6= 0, dua (∗) v^e phuong trnh tuy^en tnh c^ap m^o. t: 1 x z0 + z = 1 + x x + 1 . . C Nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at: z = x + 1 2 . . x + C1 1 2(x + 1) Bi^en thi^enhang s^o cu^oi cungduo. c: z = ⇒ y0 = = 2(x + 1) z x2 + C . . 1 Suy ra nghi^e.m t^o'ng quat cu'a phuong trnh:  2 x ln |x2 + C | + √ arctg √ + C nˆe´u C > 0  1 2 1 C1 C1√ 1 x − −C  2 √ √ 1 ´ ln |x + C1| + ln | | + C2 nˆeu C1 < 0 −C1 x + −C1 Chuy y = C laNKD . . 67) Gia'i phuong trnh: x2y0 = y(x + y) 1 1 . . HD gia’i: x2y0 = y(x + y) ⇔ y0 − = y2 : phuong trnhBernouilli y x2 1 1 D- a. t z = y−1 (y 6= 0) : −z0 − z = . . . x x2 NTQ cu'a phuong trnhthu^an nh^at: z = Cx 1 1 bi^en thi^enhang s^o C: C(x) = ε − . V^a. y z = x(ε − ) 2x2 2x2 2x V^a. y nghi^e.m t^o'ng quat la: y = εx2 − 1 . . 68) Gia'i phuong trnh: yy” − (y0)2 = y3  1 y(0) = − thoa' 2 y0(0) = 0
  16. 16 www.VNMATH.com D- at 0 00 0 thay vaophu.o.ng trnh HD gia’i: . y = p(y); y = p.py dp py − p2 = y3, dy . . . da. t ti^ep: p(y) = y.z(y) dua phuong trnh v^e dz 1 dy = ⇒ z2 = 2(y + C ) ⇔ = yp|2y + C| dy z 1 dx 1 . Do di^eu ki^e.n y(0) = − ; y0(0) = 0 ⇒ C = 1. Tu dosuy ra: 2 p dy p |2y + 1| − 1 = y |2y + 1| ⇒ ln = x + C2. dx p|2y + 1| + 1 1 do y(0) = − ⇒ C = 0. 2 2 p|2y + 1| − 1 V^a. y nghi^e.m ri^engc^an tmthoa' : ln = x. p|2y + 1| + 1 √ . . 2y x 69) Gia'i phuong trnh: ydx + 2xdy = dy thoa' di^eu ki^e.n y(0) = π cos2 y - . . . 0 2 2 1 HD gia’i: Dua phuong trnh v^e da.ng x + x = .x 2 (Bernoulli) (∗) y cos2 y - 1 0 0 1 − 1 0 Da. t z = x 2 ta co z = x + x 2 x thay vao (∗) 2 1 1 z0 + z = y cos2 y c Nghi^e.m t^o'ng quat: z = bi^en thi^enhang s^o: y y C0 = ⇒ C(y) = ytgy + ln | cos y| + ε cos2 y 1 ε V^a. y Z = tgy + ln | cos y| + y y . . 1 ε √ VaTPTQ cu'a phuong trnh: tgy + ln | cos y| + = x y y 1 √ y(0) = π ⇒ ε = 0 v^a. y TPR : tgy + ln | cos y| = x y . . 70) Gia'i phuong trnh: xydy = (y2 + x)dx . . HD gia’i: Do y = 0 kh^ongpha'i langhi^e.m, chia hai v^e cho xy bi^en d^o'i phuong trnh 1 . . . v^e da.ng: y0 − y = y−1 Bernouilli; D- a. t z = y2 dua phuong trnh v^e da.ng: x 2 z0 − z = 2 → z = −2x + Cx2 x V^a. y TPTQ: y2 = −2x + Cx2 . . √ 71) Gia'i phuong trnh: (y + xy)dx = xdy
  17. www.VNMATH.com 17 - . . . 0 1 1 1 HD gia’i: Dua phuong trnh v^e da.ng y − y = √ .y 2 ; x 6= 0 x x - 1 0 1 1 . . √ Da. t z = y 2 : z − z = √ phuong trnh tuy^en tnh gia'i ra z = x(ln x + C) 2x x V^a. y nghi^e.m t^o'ng quat: y = x(ln x + C)2 . . √ 72) Gia'i phuong trnh: xy0 − 2x2 y = 4y . . √ 1 HD gia’i: Phuong trnh Bernouilli, da. t z = y1−α = y ⇒ z0 = √ 2 y . . . 4 phuong trnh tro' thanh: z0 − z = 2x → NTQ z = Cx4 − x2 x V^a. y nghi^e.m t^o'ng quat: y = (Cx2 − 1)2x4. . . 73) Gia'i phuong trnh: 2x2y0 = y2(2xy0 − y) HD gia’i: Xem x lahamtheo bi^en y : x0y3 − 2xy2 = −2x2 Bernouilli 1 . . . 2z 2 D- a. t z = , phuong trnhtro' thanh: z0 + = → TPTQ: y2 = x ln Cy2, nghi^e.m x y y3 kydi. y = 0. . . 74) Tmnghi^e.m ri^engcu'a phuong trnh: x2y0 = y(x + y) thoa' ma~n di^eu ki^e.n d^au y(−2) = −4. . . . . . HD gia’i: Do y(−2) = −4 n^en y 6≡ 0.D- ua phuong trnhv^e phuong trnhBernouilli: y2 1 1 0 . Ti^ep tuc dat −1 du.a phu.o.ng trnh v^e PT tuy^en tnh 0 . y − 1y = 2 . . z = y z + z = − 2 x . . . . . x .x. NTQ cu'a phuong trnhthu^an nh^at tuong ung: z = Cx, bi^en thi^enhang s^o duo. c 1 . . . 2x C(x) = Cx − . Nhu v^a. y nghi^e.m cu'a phuong trnhban d^au la: y = .D- i^eu ki^e.n 2x Cx2 − 1 1 4x d^au cho C = . V^a. y nghi^e.m ri^engc^an tmla y = 2 x2 − 1 . . 75) Gia'i phuong trnh: y0 − xy = −xy3 . . . . . . HD gia’i: Phuong trnh: y0 − xy = −xy3 laphuong trnh Bernouilli, gia'i ra duo. c y2(1 + Ce−x) = 1 . . 76) Gia'i phuong trnh: xy0 + y = y2 ln x. . . . . . . HD gia’i: Phuong trnh xy0 + y = y2 ln x laphuong trnhBernouilli, gia'i ra duo. c 1 y = . 1 + Cx + ln x . . y √ 77) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 − 4 = x y x
  18. 18 www.VNMATH.com . . √ . . . HD gia’i: D- ^aylaphuong trnh Bernoulli, bang cach da. t z = y ta dua phuong 2 x trnhv^e da.ng z0 − z = vaconghi^e.m t^o'ng quat la x 2 1 z = x2( ln |x| + C). 2 . . V^a. y nghi^e.m t^o'ng quat cu'a phuong trnh la 1 y = x4( ln |x| + C)2. 2 . . y 78) Tmnghi^e.m t^o'ng quat cu'a cac phuong trnhsau: y0 + = y2xtgx. x . . HD gia’i: D- ^aylaphuong trnhBernoulli vaconghi^e.m t^o'ng quat la 1 y = . Cx + x ln | cos x| . . 79) Gia'i phuong trnh: y2dx + (2xy + 3)dy = 0 ∂P ∂Q HD gia’i: P (x, y) = y2,Q(x, y) = 2xy + 3; = = 2y ∂y ∂x (1) ⇔ d(xy2 + 3y) = 0. V^a. y xy2 + 3y = C . . 80) Gia'i phuong trnh: ex(2 + 2x − y2)dx − yexdy = 0 ∂P ∂Q . . . . . . . HD gia’i: = = −2yex suy ra phuong trnhtuong duong voi: dex(2x−y2) = ∂y ∂x 0. V^a. y ex(2x − y2) = C. . . 2 3 2 p 81) Gia'i phuong trnh: (y + 1) 2 dx + (y + 3xy 1 + y2)dy = 0 2 3 2 p ∂P ∂Q p HD gia’i: p = (y + 1) 2 ; Q = y + 3xy 1 + y2 ⇒ = = 3y 1 + y2 (∗) ∂y ∂x Suy ra nghi^e.m t^o'ng quat cu'a (∗) la: x y Z Z P (x, 0)dx + Q(x, y)dy = C 0 0 3 y 2 3 ⇔ + x(1 + y ) 2 = C 3 . . 82) Gia'i phuong trnh: (y cos2 x − sin x)dy = y cos x(y sin x + 1)dx ∂P ∂Q HD gia’i: = = y sin 2x + cos x ∂y ∂x
  19. www.VNMATH.com 19 NTQ: x y 2 R R y 2 P (x, y0)dx + Q(x, y)dy = C ⇔ y sin x − cos x = C x0=0 y0=0 2 . . 83) Gia'i phuong trnh: (2x + 3x2y)dx = (3y2 − x3)dy . . HD gia’i: Phuong trnh vi ph^antoanph^an: x2 + x3y − y3 = C . . x (x2 + 1) cos y 84) Gia'i phuong trnh: ( + 2)dx − dy = 0 sin y 2 sin2 y ∂P ∂Q x cos y HD gia’i: = = − ∂y ∂x sin2 y TPTQ: x y Z π Z x2 (x2 + 1) 1 P (x, )dx + Q(x, y)dy = C ⇔ + 2x − ( − 1) = C 2 2 2 sin y 0 π 2 . . 85) Gia'i phuong trnh: (y + ex sin y)dx + (x + ex cos y)dy = 0 . . HD gia’i: Phuong trnh vi ph^antoanph^an, nghi^e.m t^o'ng quat: xy + ex sin y = C. . . 86) Gia'i phuong trnh: (x + sin y)dx + (x cos y + sin y)dy = 0 . . HD gia’i: Phuong trnhvi ph^antoanph^an: NTQ x2 + 2(x sin y − cos y) = C. . . x3 87) Gia'i phuong trnh: 3x2(1 + ln y)dx = (2y − )dy y . . HD gia’i: Phuong trnhvi ph^antoanph^an: Nghi^e.m t^o'ng quat: x3(1 + ln y) − y2 = C . . x3 88) Tmnghi^e.m t^o'ng quat cu'a phuong trnhvi ph^an: 3x2(1 + ln y)dx = (2y − )dy y . . HD gia’i: D- ^aylaphuong trnhvi ph^antoanph^an cotch ph^ant^o'ng quat la: x3(1 + ln y) − y2 = C . . 89) Ha~y tmnghi^e.m t^o'ng quat cu'a phuong trnh: (x + sin y)dx + (x cos y + sin y)dy = 0 HD gia’i: PTVPTP cotch ph^ant^o'ng quat: x2 + 2(x sin y − cos y) = C
  20. 20 www.VNMATH.com . . 90) Ha~y tmnghi^e.m t^o'ng quat cu'a phuong trnh:  1 y2   x2 1 − dx + − dy = 0 x (x − y)2 (x − y)2 y x xy HD gia’i: PTVPTP cotch ph^ant^o'ng quat: ln + = C y x − y . . 91) Tmnghi^e.m t^o'ng quat cu'a phuong trnhvi ph^an: (sin xy + xy cos xy)dx + x2 cos xydy = 0 . . HD gia’i: Phuong trnhvi ph^antoanph^an conghi^e.m t^o'ng quat la x sin(xy) = C. . . . 92) Ha~y tmthua s^o tch ph^ancu'a phuong trnh: (x + y2)dx − 2xydy = 0 . . suy ra nghi^e.m t^o'ng quat cu'a phuong trnh. . . . 1 HD gia’i: Thua s^o tch ph^ancu'a phuong trnhla µ(x) = . Nh^anhai v^e cu'a x2 . . . y2 phuong trnh cho thua s^o tch ph^anr^oi gia'i ra x = Ce x . . . 93) Gia'i phuong trnh: 2xy ln ydx + (x2 + y2py2 + 1)dy = 0 . . . 1 HD gia’i: D- ^aylaphuong trnh vi ph^antoanph^an, thua s^o tch ph^an: µ(y) = nh^an y . . . . . 2 1 2 3 thua s^o tch ph^anvaohai v^e cu'a phuong trnhr^oi gia'i ra duo. c: x ln y+ (y +1) 2 = 0 3 . . 94) Tmnghi^e.m cu'a phuong trnh (x3 + xy2)dx + (x2y + y3)dy = 0. tho'a di^eu ki^e.n y(0) = 1. . . HD gia’i: D- ^aylaphuong trnhvi ph^antoanph^an NTQ la: x4 + 2x2y2 + y4 = C . tho'a di^eu ki^e.n y(0) = 1 khi C = 1. . . 95) Tmtch ph^ant^o'ng quat cu'a cac phuong trnhsau: a) − 2xydy + (y2 + x2)dx = 0 . . . 1 . . . HD gia’i: Ta tmduo. c thua s^o tch ph^an µ(x) = .D- ua phuong trnhda~ cho v^e x2 da.ng vi ph^antoanph^an. Khi donghi^e.m t^o'ng quat la x2 − y2 = Cx.
  21. www.VNMATH.com 21 . . 96) a) Chung minh rang h^e. cac vecto {e2x, e−x, cos x} lah^e. d^o.c l^a.p tuy^en tnh. . Tnh di.nh thuc Wronski cu'a chung. b) Tmtch ph^ant^o'ng quat cu'a cac phu.o.ng trnhsau: px2 − ydy − 2x(1 + px2 − y)dx = 0. HD gia’i: a) Dungdi.nh ngh~a ki^e'm tra h^e. d^o. c l^a. p tuy^en tnh. - . x D.inh thuc Wronski W [y1, y2, y3](x) = 3e (3 cos x − sin x). b) D- ^aylaphu.o.ng trnhvi ph^antoanph^an. Tch ph^ant^o'ng quat cu'a phu.o.ng trnh la 2 2 2 3 x + (x − y) 2 = C 3 . . x2 97) Tmtch ph^ant^o'ng quat cu'a cac phuong trnhsau: ( − y2)dy − 2xdx = 0. y . . . 1 . . . HD gia’i: Ta tmduo. c thua s^o tch ph^an µ(x) = .D- ua phuong trnhda~ cho v^e y da.ng vi ph^antoanph^an. Khi donghi^e.m t^o'ng quat la 2x2 + y3 = Cy. . . 98) a) Chung minh rang h^e. cac vecto {ex, e2x, x2} lah^e. d^o.c l^a.p tuy^en tnh. . . b) Tmtch ph^ant^o'ng quat cu'a phuong trnhsau: (x − y)dy + (x + y)dx = 0. HD gia’i: . . a) Ki^e'm tra h^e. phuong trnh lad^o. c l^a. p tuy^en tnh . . . y2 x2 b) D- ^aylaphuong trnh vi ph^antoanph^an n^enta co d(xy − + ) = 0. 2 2 V^a. y tch ph^ant^o'ng quat la x2 − y2 + 2xy = C. . . 99) a) Chung minh rang h^e. cac vecto {1, x, ex} lah^e. d^o.c l^a.p tuy^en tnh. . . b) Tmtch ph^ant^o'ng quat cu'a phuong trnhsau: (x2 − y)dx + xdy = 0 HD gia’i: a) Dungdi.nh ngh~a ki^e'm tra h^e. d^o. c l^a. p tuy^en tnh . . . . 1 . . . . . b) Tmthua s^o tch ph^an,ta duo. c µ(x) = . Phuong trnhda~ cho dua duo. c v^e . . x2 da.ng phuong trnh vi ph^antoanph^an y 1 (1 − )dx + dy = 0. x2 x . . . . Gia'i phuong trnhnayta duo. c y = Cx − x2. . . 100) a) Chung minh rang h^e. cac vecto {e2x, ex, x} lah^e. d^o.c l^a.p tuy^en tnh. . . b) Tmtch ph^ant^o'ng quat cu'a phuong trnhsau: (x − y)dx − (x + y)dy = 0. HD gia’i:
  22. 22 www.VNMATH.com . . a) Ki^e'm tra h^e. phuong trnh lad^o. c l^a. p tuy^en tnh. . . b) D- ^aylaphuong trnh vi ph^antoanph^an. Suy ra tch ph^ant^o'ng quat coda.ng: x2 + y2 − 2xy = C.
  23. www.VNMATH.com 1 . . BAI` TAˆ. P PHUONG TR`INH VI PHANˆ (tiˆe´p theo) . . 101) Gia'i phuong trnh: y” + y0 = x + e−x . . . HD gia’i: Phuong trnhda. c trung λ2 + λ = 0 ⇔ λ = 0; λ = −1 . . 1 2 Nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at: y = C + C e−x . . 1 2 . . . Tmnghi^e.m ri^engduoi da.ng y = y + y , trong do y , y lacac nghi^e.m tuong ung . . 1 2 1 2 cu'a cac phuong trnh: y” + y0 = x va y” + y0 = e−x ' . . . • V λ1 = 0 langhi^e.m cua phuong trnh da. c trung n^en y1 = x(Ax + B) . . . . 1 Bang phuong phap h^e. s^o b^at di.nh duo. c: y = x2 − x 1 2 ' . . . −x • λ2 = −1 langhi^e.m cua phuong trnh da. c trung n^en: y2 = Axe   −x Thay vaovadungh^e. s^o b^at di.nh suy ra: y2 = −xe 1 Cu^oi cungNTQ: y = C + C e−x + x2 − x − xe−x 1 2 2 . . 102) Gia'i phuong trnh: 2y” + 5y0 = 29x sin x . . . 5 HD gia’i: Phuong trnh da. c trung: 2λ2 + 5λ = 0 ⇔ λ = 0, λ = − 1 2 2 5x . . − Nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at y = C + C e 2 . . . 1 2 V ±i kh^ongpha'i langhi^e.m cu'a phuong trnhda. c trung n^entmnghi^e.m ri^engda.ng: y = (Ax + B) sin x + (Cx + D) cos x . . . . 185 16 Thay vaophuong trnhduo. c: A = −2; B = ; C = −5; D = − 29 29 . . 103) Gia'i phuong trnh: y” − 2y0 + 5y = x sin 3x . . . HD gia’i: Phuong trnhda. c trung: λ2 − 2λ + 5 = 0 ⇔ λ = 1 − 2i; λ = 1 + 2i . . 1 2 NTQ cu'a phuong trnhthu^an nh^at: y = ex(C cos 2x + C sin 2x) . . 1 . 2 Do ±3i kh^ongpha'i langhi^e.m cu'a phuong trnhda. c trung n^ennghi^e.m ri^engcu'a (2) . . . . duo. c tmduoi da.ng: y = (Ax + B) cos 3x + (Cx + D) sin 3x . . 3 57 1 41 Thay vao(2) ta duo. c: A = ; B = ; C = − ; D = 26 26 13 13 . . 104) Gia'i phuong trnh: y” − 2y0 − 3y = xe4x + x2 . . . 2 HD gia’i: Phuong trnh da. c trung: λ − 2λ − 3 = 0 ⇔ λ1 = −1; λ2 = 3. ' . .   −x 3x NTQ cua phuong trnhthu^an nh^at: y = C1e + C2e . ' 0 4x Tmnghi^e.m ri^engda.ng y = y1 + y2 voi y1 langhi^e.m cua y” − 2y − 3y = xe x 6 y = e4x(Ax + B) = e4x −  1 5 25 ' 0 2 con y2 langhi^e.m ri^engcua y” − 2y − 3y = x coda.ng: 2 4 14 y = A x2 + B x + C = − x2 + x − . 2 1 1 1 3 9 27 e4x 6 1 4 14 V^a. y nghi^e.m t^o'ng quat: y = C e−x + C e3x + (x − ) − (x2 − x + ) 1 2 5 5 3 3 9
  24. 2 www.VNMATH.com . . 105) Gia'i phuong trnh: x2y” − 2y = x3 cos x  ' . .   2 bi^et m^o.t nghi^e.m cua phuong trnhthu^an nh^at la y1 = x 2 HD gia’i: Chia 2 v^e cho x2 (x 6= 0): y” − y = x cos x. . . . x2 Tmnghi^e.m ri^engthu hai cu'a phuong trnh thu^an nh^at da.ng: 2 p(x) = 0; q(x) = − . x2 Z Z 1 − R p(x)dx 2 dx 1 y2 = y1 2 e dx = x 4 = − y1 x 3x . . 1 V^a. y nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at la: y = C x2 − C . 1 2 3x ' . .    Coi C1,C2 lahamcua x, ap du.ng phuong phap hang s^o bi^en thi^en:  1 C0 x2 + C0 (− ) = 0  1 2 3x 1 C0 2x + C0 ( ) = x cos x  1 2 3x2  cos x sin x C0 = ⇒ C = + K Gia'i ra: 1 3 1 3 1 0 3 3 2 C2 = x cos x ⇒ C2 = x sin x + 3x cos x − 6x sin x + 6 cos x + K2 x2 sin x 1 K V^a. y NTQ: y = − (x3 sin x + 3x2 cos x − 6x sin x + 6 cos x) + K x2 − 2 . 3 3x 1 3x . . 2 cotgx 106) Gia'i phuong trnh: y” + y0 + y = x x . . sin x bi^et m^o.t nghi^e.m cu'a phuong trnhthu^an nh^at la y = 1 x x cotgx . HD gia’i: p(x) = , q(x) = 1, f(x) = . Tmnghi^e.m ri^engthu hai: 2 x 2 1 R sin x x R 2 sin x dx cos x R − p(x)dx R − x dx R y2 = y1 2 e dx = 2 e dx = 2 = − y1 x sin x x sin x x . . sin x cos x NTQ cu'a phuong trnhthu^an nh^at: y = C − C 1 x 2 x  sin x cos x C0 + C0 ( ) = 0 Bi^en thi^enhang s^o:  1 x 2 x x cos x − sin x x sin x + cos x cotgx C0 + C0 =  1 x2 2 x2 x cos2 x Z cos2 x Z 1 − sin2 x ⇒ C0 = ⇒ C (x) = dx + K = dx + K 1 sin x 1 sin x 1 sin x 1 Z dx Z x = − sin xdx + K = ln |tg | + cos x + K sin x 1 2 1 0 C2 = cos x → C2 = sin x + K2 V^a. y nghi^e.m t^o'ng quat: y = ··· . . ex 107) Gia'i phuong trnh: y” − 2y0 + y = 1 + x
  25. www.VNMATH.com 3 . . . HD gia’i: Phuong trnh da. c trung: λ2 − 2λ + 1 = 0 ⇔ λ = 1 . . NTQ cu'a phuong trnhthu^an nh^at: y = ex(C x + C ) . . 1 2 Dungphuong phap bi^en thi^enhang s^o tmnghi^e.m ri^engda.ng: x x y = α1(x).xe + α2(x).e .  0 x 0 x  1 α1(x).xe + α2(x).e = 0 α0 = e−x + ex ⇔ 1 x α0 (x)(ex + xex) + α0 (x).ex = 1 + 0 x  1 2 x α2 = −(xe + 1) V^a. y ( −x α1 = −e + ln |x| −x −x α2 = xe + e − x . Nhu v^a. y nghi^e.m ri^eng: y = (ln |x| − e−x)xex + (xe−x + e−x − x)ex ' x x x Vanghi^e.m t^ong quat: y = e (C1x + C2) + xe ln |x| − xe + 1 . . 108) Gia'i phuong trnh: y” + y0 = xe−x . . . 2 HD gia’i: Phuong trnhda. c trung: λ + λ = 0 ⇔ λ1 = 0; λ2 = −1 ' ' . .   −x Nghi^e.m t^ong quat cua phuong trnh thu^an nh^at: y = C1 + C2e Tmnghi^e.m ri^engda.ng: y = xe−x(Ax + B) x2 K^et qua': y = C + C e−x − ( + x)e−x 1 2 2 . . 109) Gia'i phuong trnh: y” − 4y0 + 5y = e2x + cos x . . . 2 HD gia’i: Phuong trnh da. c trung: λ − 4λ + 5 = 0 ⇔ λ1 = 2 − i; λ2 = 2 + i ' 2x Nghi^e.m t^ong quat: y = e (C1 cos x + C2 sin x) . 2x Tmnghi^e.m ri^engda.ng: y = y1 + y2 voi y1 = Ae ; y2 = A cos x + B sin y ⇒ y1 = 1 1 e2x; y = cos x − sin x 2 8 8 1 Nghi^e.m t^o'ng quat: y = e2x(C cos x + C sin x) + e2x + (cos x − sin x) 1 2 8 . . 110) Gia'i phuong trnh: y” + 4y0 + 4y = 1 + e−2x ln x . . . HD gia’i: Phuong trnh da. c trung: λ2 + 4λ + 4 = 0 ⇔ λ = −2 −2x NTQ : y = e (C1x + C2) −2x −2x Tmnghi^e.m ri^engda.ng: y = α1(x).xe + α2e . ( 0 −2x 0 −2x α1(x).xe + α2e = 0 0 −2x −2x 0 −2x −2x α1(e − 2xe ) + α2(−2e ) = 1 + e ln x  1 α0 = e−2x + ln x → α = e−2x + x ln |x| − x  1 1 2 1 x2 1 x2 α0 = −x(e−2x + ln x) → α = e2x + − xe2x − ln x  2 2 4 4 2 2 ⇒ nghi^e.m ri^eng ⇒ nghi^e.m t^o'ng quat: 1 3x2 x2 y = e−2x(C x + C ) + e−2x( e2x − + ln x) 1 2 4 4 2 . . 111) Gia'i phuong trnh: y” + y0 = e−x(sin x − cos x)
  26. 4 www.VNMATH.com . . . . HD gia’i: D- a. t y = e−xz thay vaophuong trnh duo. c: z” − z0 = sin x − cos x. . . . Phuong trnh da. c trung: λ2 − λ = 0 ⇔ λ = 0, λ = 1 ' x Nghi^e.m t^ong quat: z = C1 + C2e . Tmnghi^e.m ri^engda.ng: z = A cos x + B sin x ⇒ A = 1,B = 0. ' −x x V^a. y nghi^e.m t^ong quat la: y = e (C1 + C2e + cos x) . . 112) Gia'i phuong trnh: y” − 4y0 + 8y = e2x + sin 2x . . . 2 HD gia’i: Phuong trnhda. c trung: λ − 4λ + 8 = 0 ⇔ λ1 = 2 − 2i; λ2 = 2 + 2i ' . .   2x Nghi^e.m cua phuong trnh thu^an nh^at: y = e (C1 cos 2x + C2 sin 2x) . ' 0 2x Nghi^e.m ri^engda.ng y = y1 + y2 voi y1 langhi^e.m ri^engcua y” − 4y + 8y = e 1 da.ng y = Ae2x → A = ; y langhi^e.m ri^engcu'a y” − 4y0 + 8y = sin 2x 1 4 2 1 1 da.ng y = A cos 2x + B sin 2x → A = ,B = . 2 10 20 V^a. y nghi^e.m t^o'ng quat: 1 1 y = e2x(C cos 2x + C sin 2x) + e2x − (2 cos 2x + sin 2x) 1 2 4 20 . . . 1 113) Gia'i phuong trnh: y” + y = sin x . . . HD gia’i: Phuong trnhda. c trung: λ2 + 1 = 0 ⇔ λ = ±i NTQ : y = C1 cos x + C2 sin x Tmnghi^e.m ri^engda.ng: y = α1(x) cos x + α2(x) sin x Bang cach bi^en thi^enhang s^o  α0 cos x + α0 sin x = 0 ( 0 (  1 2 α1 = −1 α1 = −x 1 ⇒ cos x ⇒ α0 (− sin x) + α0 cos x = α0 = α = ln sin x  1 2 sin x 2 sin x 2 ' V^a. y nghi^e.m t^ong quat: y = C1 cos x + C2 sin x − x cos x + sin x ln sin x . . x 114) Gia'i phuong trnh: y” − 3y0 + 2y = 2x2 − 5 + 2ex cos 2 2 HD gia’i: λ − 3λ + 2 = 0 ⇔ λ1 = 1; λ2 = 2 x 2x NTQ: y = C1e + C2e x 2x     Tmnghi^e.m ri^engda.ng: y = α1(x)e + α2(x)e bang cach bi^en thi^enhang s^o: ( 0 x 0 2x α1e + α2e = 0 x α0 ex + α0 (2e2x) = 2x2 − 5 + 2ex cos 1 2 2  0 −x 2 x α1 = −e (2x − 5) − 2 cos 2 x α0 = e−2x(2x2 − 5) + 2e−x cos  2 2  −x 2 x α1 = e (2x − 4x − 1) − 4 sin ⇒ 2 1 −2x 2 −2x 1 −2x 8 −2x x 1 −x x α2 = − [e (2x − 5) + 2(xe + e )] + (−e cos + e sin ) 2 2 3 2 2 2 . . . Tu doconghi^e.m t^o'ng quat cu'a phuong trnh.
  27. www.VNMATH.com 5 . . 115) Gia'i phuong trnh: y” − 4y = (2 − 4x)e2x ' −2x 2x HD gia’i: Nghi^e.m t^ong quat: y = C1e + C2e 2 2 Nghi^e.m ri^engda.ng: y = xe2x(Ax + B); A = − ,B = 3 3 2 → Nghi^e.m t^o'ng quat: y = C e−2x + C e2x + xe2x(1 − x) 1 2 3 . . ex 116) Gia'i phuong trnh: y” − 2y0 + y = + cos x x ' x HD gia’i: Nghi^e.m t^ong quat: y = e (C1x + C2) ∗ x x    nghi^e.m ri^engda.ng: y = α1xe + α2e bi^en thi^enhang s^o:   1 1 −x  0 −x α1 = ln |x| + e (sin x − cos x) α1 = e cos x 2 x → 1 α0 = −(1 + xe−x cos x) α = −x − (xe−x(sin x − cos x) + e−x sin x) 2  2 2 ⇒ Nghi^e.m t^o'ng quat . . 117) Gia'i phuong trnh: y” − 2y0 + 2y = x(ex + 1) . . . HD gia’i: Phuong trnh da. c trung: λ2 − 2λ + 2 = 0 ⇔ λ = 1 − i λ = 1 + i . . . . 1 . 2 Nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at tuong ung: x y = e (C1 cos x + C2 sin x) . ' 0 x Nghi^e.m ri^engda.ng y = y1 + y2 voi y1 langhi^e.m ri^engcua y” − 2y + 2y = xe x ' 0 coda.ng y1 = e (Ax + B) → A = 1,B = 0; Va y2 langhi^e.m ri^engcua y” − 2y + 2y = x 1 y2 = Ax + B → A = B = . 2 . . V^a. y nghi^e.m t^o'ng quat cu'a phuong trnh: 1 y = ex(C cos x + C sin x) + xex + (x + 1) 1 2 2 . . . e−x 118) Gia'i phuong trnh: y” + 2y0 + y = sin x + x . . . HD gia’i: Phuong trnh da. c trung λ2 + 2λ + 1 = 0 ⇔ λ = −1 (b^o. i 2) ' −x Nghi^e.m t^ong quat: y = e (C1x + C2). −x −x Tmnghi^e.m ri^engda.ng: y = α1(x)xe + α2(x)xe Bi^en thi^enhang s^o:   ex 0 x 1 α = e sin x + α1 = (sin x − cos x) + ln |x|  1  2 x ⇒ x x 2 0 x x xe e x α2 = −xe sin x − α = −[ (sin x − cos x) + cos x] − 2  2 2 2 4 cos x x2e−x Suy ra nghi^e.m t^o'ng quat: y = e−x(C x + C ) + xe−x ln |x| − − . 1 2 2 4
  28. 6 www.VNMATH.com . . 1 119) Gia'i phuong trnh: y” + y = sin x . . . HD gia’i: Phuong trnhda. c trung λ2 + 1 = 0 ⇔ λ = ±i ' Nghi^e.m t^ong quat: y = A1 cos x + A2 sin x. ( 0 ( Bi^en thi^enhang s^o: A1 = −1 A1 = −x 0 ⇒ A2 = cotgx A2 = ln | sin x|. ' V^a. y nghi^e.m t^ong quat: y = (C1 − x) cos x + (C2 + ln | sin x|) sin x. . . 120) Gia'i phuong trnh: y” + y = xex + 2e−x . . . HD gia’i: Phuong trnh da. c trung λ2 + 1 = 0 ⇔ λ = ±i ' Nghi^e.m t^ong quat: y = C1 cos x + C2 sin x. Tmnghi^e.m ri^engda.ng:   1 2A = 1 A =   2 x −x 1 y = (Ax + B)e + Ce → A + B = 0 → B = − 2C = 2  2 C = 1 1 V^a. y nghi^e.m t^o'ng quat: y = C cos x + C sin x + (x − 1)ex + e−x 1 2 2 . . 121) Gia'i phuong trnh: y” − y0 − 2y = cos x − 3 sin x . . . 2 HD gia’i: Phuong trnhda. c trung λ + λ − 2 = 0 ⇔ λ1 = −2; λ2 = 1 ' −2x x Nghi^e.m t^ong quat: y = C1e + C2e Tmnghi^e.m ri^engda.ng: ( ( B − 3A = 1 A = 0 y = A cos x + B sin x → → −A − 3B = −3 B = 1 ' −2x x V^a. y nghi^e.m t^ong quat: y = C1e + C2e + sin x . . 122) Gia'i phuong trnh: y” − 2y0 = 2 cos2 x . . . 2 HD gia’i: Phuong trnh da. c trung λ − 2λ = 0 ⇔ λ1 = 0; λ2 = 2 ' 2x Nghi^e.m t^ong quat: y = C1 + C2e . Tmnghi^e.m ri^engda.ng: y = Ax + B cos 2x + C sin 2x  1  A = − −2A = 1  2 . .   1 Thay vaoduo. c: −4(B + C) = 1 → B = −   8 4(B − C) = 0  1 C = − 8 x 1 V^a. y nghi^e.m t^o'ng quat: y = C + C e2x − − (cos 2x + sin 2x) 1 2 2 8 . . 123) Gia'i phuong trnh: y” + y = sin x + cos 2x
  29. www.VNMATH.com 7 . . . HD gia’i: Phuong trnh da. c trung λ2 + 1 = 0 ⇔ λ = ±i ' ' . .   Nghi^e.m t^ong quat cua phuong trnh thu^an nh^at: y = C1 cos x + C2 sin x. Tmnghi^e.m ri^engda.ng: y = x(A cos x + B sin x) + C cos 2x + D sin 2x . . . . 1 1 Thay vaophuong trnhvad^ong nh^at duo. c: A = − ; B = 0; C = − ; D = 0 2 3 1 1 V^a. y nghi^e.m t^o'ng quat: y = C cos x + C sin x − x cos x − cos 2x. 1 2 2 3 . . 124) Tmnghi^e.m t^o'ng quat cu'a phuong trnhvi ph^an: y00 − 2y0 = 2 cos2 x HD gia’i: . . . 2 ' ' Phuong trnhda. c trung λ − 2λ = 0 ⇐⇒ λ1 = 0; λ2 = 2. Nghi^e.m t^ong quat cua . .    . . . 2x phuong trnhtuy^en tnh thu^an nh^at tuong ung: y = C1 + C2e . Tmnghi^e.m ri^eng da.ng: y∗ = Ax + B cos 2x + C sin 2x . . 1 1 1 D- uo. c A = − ; B = − ; C = − . V^a. y NTQ: 2 8 8 x 1 y = C + C e2x − − (cos 2x + sin 2x) 1 2 2 8 . . 125) Tmnghi^e.m t^o'ng quat cu'a phuong trnhvi ph^an: x x x (x + e y )dx + e y (1 − )dy = 0. y 2 . . x x HD gia’i: Phuong trnhvi ph^antoanph^an cotch ph^ant^o'ng quat; + ye y = C. 2 . . 126) Gia'i phuong trnh: y00 − 6y0 + 9y = 25ex sin x. . . . . . HD gia’i: NTQ cu'a phuong trnhtuy^en tnh thu^an nh^at tuong ung y = (C +C x)e3x. . . 1 2 Tmnghi^e.m ri^engda.ng: y∗ = ex(A cos x + B sin x); duo. c A = 4; B = 3. V^a. y NTQ: 3x x y = (C1 + C2x)e + e (3 cos x + 4 sin x) . . 127) Ha~y tmnghi^e.m t^o'ng quat cu'a phuong trnh: y00 − 2y0 + 2y = x(ex + 1) . . . 2 ' HD gia’i: Phuong trnhda. c trung λ − 2λ + 2 = 0 ⇐⇒ λ1 = 1 ± i. Nghi^e.m t^ong quat ' . .    . . . x cua phuong trnhtuy^en tnh thu^an nh^at tuong ung: y = e (C1 cos x + C2 sin x). Tm ∗ . ' 00 0 x nghi^e.m ri^engda.ng: y = y1 + y2; voi y1 langhi^e.m ri^engcua y − 2y + 2y = xe , coda.ng x ' 00 0 y1 = e (Ax + B) =⇒ A = 1; B = 0 va y2 langhi^e.m ri^engcua y − 2y + 2y = x, coda.ng 1 y = A0x + B0 =⇒ A0 = B0 = . v^a. y nghi^e.m t^o'ng quat: 2 2 1 y = ex(C cos x + C sin x) + xex + (x + 1) 1 2 2
  30. 8 www.VNMATH.com . . 128) Tmnghi^e.m t^o'ng quat cu'a phuong trnh: x2y00 − 2y = x3 cos x  ' . .   2 bi^et m^o.t nghi^e.m ri^engcua phuong trnhthu^an nh^at la y1 = x . . . 1 . HD gia’i: TmNR da.ng y = uy = ux2 duo. c y = − . Nhu v^a. y NTQ: y = 2 1 2 3x C2 . . 1 C x2 + . Bi^en thi^enhang s^o duo. c C0 = − cos x; C0 = x3 cos x 1 x 1 3 2 . . 129) Gia'i phuong trnhvi ph^ansau d^ayn^eu bi^et m^o.t nghi^e.m ri^engcu'a no . coda.ng dathuc: (x2 + 1)y00 − 2y = 0 ~  2 ' . . HD gia’i: D^e th^ay y1 = x + 1 lam^o. t nghi^e.m ri^engcua phuuong trnh,nghi^e.m ri^eng .  . thu hai d^o. c l^a. p tuy^en tnh voi y1 la: Z Z 1 − R 0.dx 2 dx 1 2 x y2 = y1 2 e dx = (x + 1) 2 2 = (x + 1)( 2 + arctan x) y1 (x + 1) 2 x + 1 x V^a. y NTQ: y = C (x2 + 1) + C (x2 + 1)( + arctan x) 1 2 x2 + 1 . . 130) Tmnghi^e.m t^o'ng quat cu'a phuong trnh: y00 + y = sin x + cos 2x. . . . 2 ' ' HD gia’i: Phuong trnhda. c trung λ + 1 = 0 ⇐⇒ λ1 = ±i. Nghi^e.m t^ong quat cua . .    . . . phuong trnhtuy^en tnh thu^an nh^at tuong ung: y = C1 cos x + C2 sin x. Tmnghi^e.m . . . 1 ri^engda.ng: y∗ = y + y ; voi y langhi^e.m ri^engcu'a y00 + y = sin x, duo. c y = − x cos x 1 2 1 1 2 . . 1 va y langhi^e.m ri^engcu'a y00 + y = cos 2x, duo. c y = − cos 2x. V^a. y nghi^e.m t^o'ng quat: 2 2 3 1 1 y = C cos x + C sin x − x cos x − cos 2x 1 2 2 3 . . 131) Tmnghi^e.m t^o'ng quat cu'a phuong trnhvi ph^an: y00 + 10y0 + 25y = 4e−5x . . . 2 ' HD gia’i: Phuong trnhda. c trung r + 10r + 25 = 0 giai ra r1 = r2 = 5 ' . .   −5x ' . . NTQ cua phuong trnh thu^an nh^at: y = (C1 + C2x)e va NR cua phuong trnh kh^ongthu^an nh^at: y∗ = 2x2e−5x. V^a. y −5x 2 −5x NTQ: y = (C1 + C2x)e + 2x e . . sin x 132) Bi^et rang phuong trnh xy00 + 2y0 + xy = 0 conghi^e.m ri^engda.ng y = . . . x Ha~y tmnghi^e.m t^o'ng quat cu'a phuong trnh. . sin x cos x HD gia’i: Nghi^e.m ri^engd^o. c l^a. p tuy^en tnh voi y = la y = . V^a. y nghi^e.m x x t^o'ng quat cu'a phu.o.ng trnh la sin x cos x y = C . + C . 1 x 2 x
  31. www.VNMATH.com 9 . . 133) Tmnghi^e.m t^o'ng quat cu'a phuong trnhvi ph^an: y00 + y0 = 4x2ex ' ' . .   . . . −x HD gia’i: Nghi^e.m t^ong quat cua phuong trnhthu^an nh^at tuong ung: y = C1+C2e ∗ 2 −x ' Tmnghi^e.m ri^engda.ng: y = (A1x + A2x + A3)e , giai ra A1 = 2; A2 = −6; A3 = 7. . . 134) Tmnghi^e.m t^o'ng quat cu'a phuong trnhvi ph^an: y00 + 3y0 + 2y = x sin x ' ' . .   . . . −x HD gia’i: Nghi^e.m t^ong quat cua phuong trnhthu^an nh^at tuong ung: y = C1e + −2x ' . .   . . . . C2e . Nghi^e.m ri^engcua phuong trnhkh^ongthu^an nh^at duo. c tmduoi da.ng: y = . . 3 17 1 3 (A x + A ) cos x + (B x + B ) sin x vatmduo. c A = − ; A = ; B = ; B = . 1 2 1 2 1 10 2 50 1 10 2 25 . . . 135) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − 2y0 + 2y = xex . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 y = ex(C cos x + C sin x) + (x + 1) + ex 1 2 2 . . . 136) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” + y = cos 2x. . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 1 y = C cos x + C sin x + − cos 2x. 1 2 2 6 . . 137) Tmnghi^e.m t^o'ng quat cu'a phuong trnh: (1 − x2)y” − 2xy0 + 2y = 0  khi bi^et m^o.t nghi^e.m ri^eng y1 = x. . 2x HD gia’i: Chuy^e'n v^e da.ng y” + p (x)y0 + p (x)y = 0. Voi p (x) = − n^ennghi^e.m 1 2 1 1 − x2 t^o'ng quat cu'a phu.o.ng trnh da~ cho la R 2x dx Z e 1−x2 Z dx y = x{ C dx + C } = x{C + C } 1 x2 2 1 x2(1 − x2) 2 1 1 1 + x x 1 + x = x{(− + ln ) + C } = C x + C ( ln − 1). x 2 1 − x 2 2 1 2 1 − x . . . 138) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − 3y0 + 2y = 2 + ex . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la:
  32. 10 www.VNMATH.com x 2x x y = C1e + C2e − 2xe . . . . 139) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − y0 = sin2 x. . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 1 y = C + C ex + cos x − ln x. 1 2 2 2 . . . 140) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − 2y0 + 10y = xex . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 y = C ex cos 3x + C ex sin 3x − xex. 1 2 9 . . . 141) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” + y = cos 2x + sin x. . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 1 y = C ex cos x + C ex sin x − cos 2x − x cos x. 1 2 3 2 . . . 142) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − 2y0 + y = xex . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: x3 y = C ex + C xex + ex. 1 2 6 . . . 143) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” + y = cos 2x. . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 1 y = C + C e−x + sin 2x − cos 2x. 1 2 10 5 . . 3 1 144) Tmnghi^e.m t^o'ng quat cu'a phuong trnh y” + y0 + y = 0, x x2 1 khi bi^et m^o.t nghi^e.m ri^engcoda.ng y = . 1 x
  33. www.VNMATH.com 11 . . . . . . . . . HD gia’i: Phuong trnhda~ cho tuong duong voi phuong trnh x2y” + 3xy0 + y = 0. . . . . . . D- ^aylaphuong trnhEuler n^enta coth^e' dua v^e phuong trnhtuy^en tnh voi h^e. s^o hang bang cach dat t Khi dophu.o.ng trnh da~ cho tro'. thanh 0 . x = e . yt” + 2yt + y = 0. . . −t −t ' . . Phuong trnhnayconghi^e.m la y = C1e + C2te . V^a. y nghi^e.m cua phuong trnh C ln |x| da~ cho la y = 1 + C . x 2 x . . . 145) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang a) y” − 3y0 + 2y = 2e2x . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: x 2x 2x y = C1e + C2e + 2e . . . . 146) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang 1 a) y” + y = cos2 x ' . .   . . HD gia’i: Nghi^e.m cua phuong trnhthu^an nh^at y = C1 cos x + C2 sin x. Dungphuong sin x 1 phap bi^en thi^enhang s^o ta du.o.c 0 va 0 . C1(x) = − 2 C2(x) = . . . cos x cos x V^a. y nghi^e.m cu'a phuong trnh la sin x 1 + sin x y = C cos x + C sin x − 1 + ln | |. 1 2 2 1 − sin x . . . 147) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − 2y0 + 2y = x + ex . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 y = ex(C cos x + C sin x) + (x + 1) + ex. 1 2 2 . . . 148) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” + y = cos2 x. . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. 1 1 Nghi^e.m t^o'ng quat la: y = C cos x + C sin x + − cos 2x. 1 2 2 6 . . 1 149) Tmnghi^e.m t^o'ng quat cu'a phuong trnh xy” + y0 − y = 0, a x khi bi^et m^o.t nghi^e.m ri^engcoda.ng y = . 1 x
  34. 12 www.VNMATH.com 1 . . 1 HD gia’i: y = lam^o. t nghi^e.m cu'a phuong trnh.Ta tmnghi^e.m ri^eng y = u(x) . 1 x 2 x . . . . ' ' . . Thay vaophuong trnhta tmduo. c y2 = x. V^a. y nghi^e.m t^ong quat cua phuong trnh la C y = 1 + C x. x 2 . . . 150) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang a) y” − 3y0 + 2y = 2ex HD gia’i: . . . D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ong thu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: x 2x x y = C1e + C2e − 2xe . . . . 151) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − y = sin x. HD gia’i: . . . D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ong thu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 1 y = C + C ex + cos x − sin x. 1 2 2 2 . . 152) Tmnghi^e.m t^o'ng quat cu'a phuong trnh x2y” − 2xy0 − 4y = 0, 1 khi bi^et m^o.t nghi^e.m ri^engcoda.ng y = . 1 x 1 . . 1 HD gia’i: y = lam^o. t nghi^e.m cu'a phuong trnh.Ta tmnghi^e.m ri^eng y = u(x) . 1 x 2 x . . . . 4 ' ' . . Thay vaophuong trnhta tmduo. c y2 = x . V^a. y nghi^e.m t^ong quat cua phuong trnh la C y = 1 + C x4. x 2 . . . 153) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” + y = x + 2ex HD gia’i: . . . D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ong thu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: x y = C1 cos x + C2 sin x + x + e . . . . 154) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − y0 + y = x.
  35. www.VNMATH.com 13 HD gia’i: . . . D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ong thu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: √ √ x 3 3 y = e 2 (C cos x + C sin x) + 1 + x. 1 2 2 2 . . . 155) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” − 2y0 + y = x + ex . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 y = C ex + C xex + 2 + x2ex. 1 2 2 . . . 156) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang y” + y = sin2 x. . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 1 y = C cos x + C sin x + + cos 2x. 1 2 2 6 . . 1 157) Gia'i phuong trnhtuy^en tnh c^ap 2 sau: xy” − y0 − y = 0. x . . . . . HD gia’i: D- ^aylaphuong trnhEuler n^enta coth^e' dua v^e phuong trnhtuy^en . . . . tnh voi h^e. s^o hang bang cach da. t x = et. Khi dophuong trnhda~ cho tro' thanh y ” − 2y0 − y = 0. t . t. Phuong trnh nayconghi^e.m la √ √ (1+ 2)t (1− 2)t y = C1e + C2e . . . V^a. y nghi^e.m cu'a phuong trnh da~ cho la √ √ 1+ 2 1− 2 y = C1x + C2x . . . . 158) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang sau: y” − 3y0 + 2y = 2 cos x . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 3 y = C ex + C e2x + cos x − sin x. 1 2 5 5 . . . 159) Gia'i phuong trnhtuy^en tnh c^ap 2 voi h^e. s^o hang sau: y” − y = sin x + ex.
  36. 14 www.VNMATH.com . . . HD gia’i: D- ^aylaphuong trnhtuy^en tnh c^ap 2 kh^ongthu^an nh^at voi h^e. s^o hang. Nghi^e.m t^o'ng quat la: 1 1 y = C + C ex + xex + cos x − sin x. 1 2 2 2 z . . 160) Dungphep d^o'i ham y = d^e' gia'i phuong trnhvi ph^an: x2 x2y” + 4xy0 + (x2 + 2)y = ex z z0x − 2z z”x2 − 4z0x + 6z HD gia’i: y = ⇒ y0 = ; y” = x2 x3 x4 . . . ex Phuong trnh tro' thanh: z” + z = ex com^o. t nghi^e.m ri^eng y = . . . . . 2 Phuong trnh thu^an nh^at cophuong trnh da. c trung λ2 + 1 = 0 ⇔ λ = ±i ex V^a. y nghi^e.m t^o'ng quat: z = C cos x + C sin x + 1 2 2 cos x sin x ex V^a. y y = C + C + 1 x2 2 x2 2x2 . . 161) Gia'i phuong trnh y” cos x + y0 sin x − y cos3 x = 0 bang phep bi^en d^o'i t = sin x 0 0 0 0 HD gia’i: t = sin x : yx = yt.tx = yt cos x 2 0 y”xx = y”tt cos x − yt sin x . . t −t sin x − sin x Thay vaophuong trnh: y”tt − y = 0 → y = C1e + C2e = C1e + C2e . . x x x 162) Tmnghi^e.m ri^engcu'a phuong trnh (x + e y )dx + e y (1 − ) = 0 y thoa' di^eu ki^e.n y(0) = 2 2 ∂P ∂Q x x x x HD gia’i: = = − e y , y 6= 0 TPTQ: + ye y = C ∂y ∂x y2 2 y(0) = 2 ⇒ C = 2. . . 163) Gia'i phuong trnhvi ph^an y” + y0tgx − y cos2 x = 0 bang phep bi^en d^o'i t = sin x . . . HD gia’i: Tuong tu. bai2 .  1 1  164) Cho bi^e'u thuc: h(x) ( − ln(x + y))dx + dy . x.+ y . x + y Ha~y tmhams^o h(x) sao cho bi^e'u thuc tr^entro' thanhvi ph^antoanph^an cu'a m^o.t ham F (x, y) vatmhams^o do.  1  HD gia’i: D- a. t P = h(x) ln (x + y) x + y 1 Q = h(x). x + y (D- i^eu ki^e.n x+y > 0) d^e' P dx + Qdy lavi ph^antoanph^an: ∂P ∂Q −h(x)(x + y + 1) h0(x)(x + y) − h(x) = ⇔ = ∂y ∂x (x + y)2 (x + y)2
  37. www.VNMATH.com 15 ⇔ h0(x + y) + h(x + y) = 0 ⇔ h0 + h = 0 ⇔ h(x) = e−x Va F (x, y) = e−x ln(x + y) . . 165) Gia'i phuong trnhvi ph^an: xy” + 2(1 − x)y0 + (x − 2)y = e−x bang phep d^o'i ^a'n ham z = yx z z0x − z . . . HD gia’i: z = yx ⇔ y = ; y0 = = ; y” = tuong tu. bai1 x x2 166) Cho P (x, y) = ex sin y + 2m2x cos y; Q(x, y) = ex cos y + mx2 sin y. Tmm d^e' P (x, y)dx + Q(x, y)dy lavi ph^antoanph^an cu'a hams^o F (x, y) naodovatmham^ay. ∂P ∂Q HD gia’i: = ⇔ 2x sin y(m2 + m) = 0 Cho.n m = 0V m = −1. ∂y ∂x . . y 1 167) Gia'i phuong trnh x2y” + 2xy0 + = 0 bang phep bi^en d^o'i x = x2 t HD gia’i:   168) Tmham µ(x2 + y2) sao cho µ(x2 + y2) (x − y)dx + (x + y)dy lavi ph^antoanph^an cu'a m^ot ham naodo. Tmham . F (x,√ y) √ F (x, y) n^eu bi^et µ(1, 1) = 0; µ( 2, 2) = ln 2 HD gia’i: P (x, y) = h(x2 + y2)(x − y); Q(x, y) = h(x2 + y2)(x + y) ∂P ∂Q D- ^e' h(x − y)dx + h(x + y)dy lavi ph^antoanph^an ta pha'i co: = ∂y ∂x D- at 2 2 0 0 . t = x + y ⇒ ht.2y(x − y) − h = ht.2y(x + y) + h C C ⇔ −h0 (x2 + y2) = h ⇔ h0 t = h ⇒ h = 1 ⇒ h = 1 . t t t x2 + y2 x y R x − 0 R x + y y C1 2 2 ⇒ F (x, y) = C1 2 2 dx + C1 2 2 dy = C1arctg + ln(x + y ) + C2 1 x + 0 0 x + y 2 2 √ √ π F (1, 1) = 0; F ( 2, 2) = ln 2 Cho: C = 2; C = −( + ln 2) 1 2 2 . . 169) Gia'i phuong trnh x2y” + xy0 + y = x bang phep d^o'i bi^en x = et 1 1 HD gia’i: x = et ta co: y0 = y0. ; y” = (y” − y0) x t x xx tt t x2 . . t Thay vaophuong trnh: y”tt + y = e ' ' . .   Nghi^e.m t^ong quat cua phuong trnh thu^an nh^at: y = C1 cos t + C2 sin t 1 Tmnghi^e.m ri^engda.ng: y + Aet; A = 2 x V^a. y nghi^e.m t^o'ng quat: y = C cos (ln x) + C sin (ln x) + 1 2 2
  38. 16 www.VNMATH.com . . 170) Gia'i phuong trnhvi ph^an: xy” − (x + 1)y0 − 2(x − 1)y + x2 = 0 . . . . . bi^et rang phuong trnhthu^an nh^at tuong ung com^o.t nghi^e.m ri^eng y = eαx . 1 voi α lahang s^o c^an xac di.nh. αx . .    . . HD gia’i: Thay nghi^e.m y1 = e vaophuong trnh r^oi d^ong nh^at duo. c α = 2 . . . x + 1 2(x − 1) D- ua phuong trnh v^e da.ng: y” − y0 − y = −x; x 6= 0 x x x + 1 2(x − 1) p(x) = − ; q(x) = − ; f(x) = −x x x x + 1 R dx 2x R −4x 1 −x Tmnghi^e.m ri^eng: y2 = e e e x dx = − (3x + 1)e . . . 9 Suy ra nghi^e.m t^o'ng quat cu'a phuong trnh thu^an nh^at: 2x −x y = C1e + C2(3x + 1)e Bi^en thi^enhang s^o:  1  1 C0 = − (3x + 1)e−2x C = (6x + 5)e−2x  1 9  1 36 1 → 1 C0 = ex C = ex  2 9  2 9 ⇒ NTQ. . . 171) Gia'i phuong trnhvi ph^an x2y00 − 4xy0 + 6y = 0 bang phep d^o'i bi^en x = et. 1 1 HD gia’i: x = et, ta co: y0 = y0. , y” = (y” − y0) x t x xx tt t x2 Phu.o.ng trnh tro'. thanh: 0 NTQ: 2 3 y”tt − 5yt + 6y = 0 ⇒ y = C1x + C2x . . 172) Gia'i phuong trnhvi ph^an: y” − (2ex + 1)y0 + e2xy = e3x bang phep d^o'i bi^en t = ex. D- ^o'i bi^en x 0 0 x 2x 0 x HD gia’i: t = e ⇒ yx = yt.e , y”xx = y”tt.e + yt.e Thay vaophu.o.ng trnh: 0 3 y”tt − 2yt + y = t ' ' . .   t Nghi^e.m t^ong quat cua phuong trnh thu^an nh^at: y = e (C1t + C2) Tmnghi^e.m ri^engda.ng y = At3 + Bt2 + Ct + D → y = t3 + 6t2 + 18t + 24 K^et qua' ex x 3x 2x x y = e (C1e + C2) + e + 6e + 18e + 24. . . 173) Gia'i phuong trnhvi ph^an: (x − 1)y” − xy0 + y = (x − 1)2e2x . . . . . bi^et m^o.t nghi^e.m ri^engcu'a phuong trnhthu^an nh^at tuong ung coda.ng y = eαx (α c^an xac di.nh). . . . x 1 HD gia’i: D- ua phuong trnh v^e: y” − .y0 + .y = (x − 1)e2x x − 1 x − 1 . x 1 Voi p(x) = ; q(x) = ; f(x) = (x − 1)e2x x − 1 x − 1 αx . .   . . .    Thay y1 = e vaophuong trnhthu^an nh^at tuong ung r^oi d^ong nh^at suy ra α = 1 R x x R −2x dx Tmnghi^e.m ri^eng y2 = e e e x−1 dx = −x
  39. www.VNMATH.com 17 x ⇒ NTQ: y = C1e + C2(−x) Bi^en thi^enhang s^o: (  x x C0 = xex C1 = xe − e + K1 1 → 1 C0 = e2x C = e2x + K 2  2 2 2 x V^a. y nghi^e.m t^o'ng quat: y = ( − 1)e2x + K ex − K x 2 1 2 . . 1 174) Gia'i phuong trnhvi ph^an: x2(x + 1)y” = 2y bi^et m^o.t nghi^e.m y = 1 + . 1 x . . . 2 HD gia’i: D- ua phuong trnh v^e: y00 − .y = 0; p(x) = 0; f(x) = 0. x2(x + 1) TmNR da.ng 2 1 Z x R 1 1 y = (1 + ) .e− 0dxdx = (1 + )(x − 2 ln |x + 1| − ) 2 x (x + 1)2 x 1 + x x + 1 1 = x + 1 − ln(x + 1)2 − . x x 1 1 x + 1 V^a. y nghi^e.m t^o'ng quat: y = C (1 + ) + C (x − − 1 + ln(x + 1)2 + 1). 1 x 2 x x . . 175) Gia'i phuong trnhvi ph^an (x2 + 1)y” − 2y = 0 . n^eu bi^et m^o.t nghi^e.m cu'a nocoda.ng dathuc. ~  2 ' HD gia’i: D^e th^ay y1 = x + 1 lam^o. t nghi^e.m ri^engcua (1). Z Z . 1 − R p(x)dx 2 dx Nghiˆe.m th´u hai: y2 = y1 2 .e dx = (x + 1) 2 2 y1 (x + 1) 1 x = (x2 + 1)( + arctgx) 2 x2 + 1 x V^a. y nghi^e.m t^o'ng quat: y = C (x2 + 1) + C (x2 + 1)( + arctgx). 1 2 x2 + 1 . . 176) Gia'i phuong trnhvi ph^an xy” + 2y0 − xy = ex bang phep d^o'i ham z = xy. . . HD gia’i: D- a. t z = xy ⇒ z0 = y + xy0; z00 = 2y0 + xy00. Thay vaophuong trnh: 00 x x z − z = e → NTQ z = C1 + C2e 1 Nghi^e.m ri^engda.ng: y = Axex → A = 2 z 1 1 V^a. y: y = = (C + C ex + xex) x x 1 2 2 . ∞ xn+1 177) Chung to' rang ham: f(x) = P . . n=0 n! langhi^e.m cu'a phuong trnh xf 0(x) − (x + 1)f(x) = 0.
  40. 18 www.VNMATH.com . ∞ xn+1 . HD gia’i: Dungtnh ch^at D'Alembert d^e' chung to' chu^o~i P h^o. i tu. voi mo.i x n=0 n! . ∞ xn+1 . Nhu v^a. y ham f(x) = P xac di.nh voi mo.i x. n=0 n! . . ∞ xn Hon n~ua: f(x) = x P = xex n! n=0 . ⇒ xf 0(x) − (x + 1)f(x) = x(x + 1)ex − (x + 1)xex = 0, ∀x di^eu pha'i chung minh. . . 178) Gia'i phuong trnh x(x2 + 6)y” − 4(x2 + 3)y0 + 6xy = 0 . bi^et rang noconghi^e.m da.ng dathuc. . . 2 2 HD gia’i: Ta tmnghi^e.m ri^engduoi da.ng y1 = Ax + Bx + C ⇒ y1 = x + 2 2 . 1 − R − 4(x +3) dx nghi^em ri^engthu hai: R x(x+6) . y2 = y1 2 e dx y1 x2(x2 + 6) 2x √ x = (x2 + 2) R dx = (x2 + 2)(x + + 2 2arctg√ ) (x2 + 2)2 (x2 + 2) 2 √ 2 3 2 x V^a. y NTQ: y = C1(x + 2) + C2[x + 4x + 2 2(x + 2)arctg√ ] 2 . . 179) Gia'i phuong trnh (2x + 1)y” + (2x − 1)y0 − 2y = x2 + x bi^et rang x2 + 4x − 1 x2 + 1 nocohai nghi^e.m ri^eng y = ; y = . 1 2 2 2 . . . HD gia’i: Tu hai nghi^e.m ri^eng y , y cu'a phuong trnhta suy ra nghi^e.m ri^engcu'a . . 1 2 phuong trnh thu^an nh^at la y = y − y = 2x − 1 . 1 1 2 Suy ra nghi^e.m thu hai: Z 1 R Z 1 R 2x−1 − p(x)dx − 2x+1 dx y2 = y1 2 e dx = (2x − 1) 2 e dx y1 (2x − 1) Z (2x + 1)e−x 1 (2x + 1)e−x Z e−x(1 − 2x) = 2(x − 1) dx = (2x − 1)[− + dx] (2x − 1)2 2 (2x − 1)2 2x − 1 = −e−x −x Suy ra NTQ: y = C1(2x − 1) + C2e . . x2 + 1 Vanghi^e.m t^o'ng quat cu'a phuong trnh ban d^au: y = C (2x − 1) + C e−x + 1 2 2 2 . . 180) Xac di.nh hang s^o α sao cho y = eαx lam^o.t nghi^e.m ri^engcu'a phuong trnhvi ph^an: . . y” + 4xy0 + (4x2 + 2)y = 0. Tmnghi^e.m t^o'ng quat cu'a phuong trnh^ay. . . 2 . . HD gia’i: Ta tmnghi^e.m ri^engduoi da.ng y = eαx thay vaoduo. c α = −1 vanghi^e.m −x2 ri^eng y1 = e 1 R 2 2 R 2 Nghi^em thu. hai: R − P (x)dx −x R 2x − 4xdx −x . . y2 = y1 2 e dx = e e e dx = xe y1 −x2 −x2 V^a. y NTQ: y = C1e + C2xe . dx  = 3x − y Gia'i h^e phu.o.ng trnh:  dt 181) . dy  = 4x − y  dt
  41. www.VNMATH.com 19 . . . 3 − λ −1 HD gia’i: Phuong trnh da. c trung = (λ − 1)2 = 0 ⇔ λ = 1 (b^o. i 2) 4 −1 − λ a = 3a − c  x (at + b)et . . a + b = 3b − d Tmnghi^e.m da.ng = thay vaoh^e. r^oi d^ong nh^at duo. c: y (ct + d)et c = 4a − c  c + d = 4b − d Cho a = C1, b = C2 ⇒ c = 2C1, d = 2C2 − C1 ( t V^ay nghi^em t^o'ng quat: x = (C1t + C2)e . . t y = (2C1t + 2C2 − C1)e . dx  = 2x + y Gia'i h^e phu.o.ng trnh:  dt 182) . dy  = 4y − x  dt . . . . . . HD gia’i: Tuong tu. bai1), phuong trnhda. c trung conghi^e.m λ = 3 (b^o. i 2) (at + b)e3t Tmnghi^e.m da.ng ⇒ a = C , c = C , b = C , d = C + C (ct + d)e3t 1 1 2 1 2 ( 3t V^ay NTQ: x = (C1t + C2)e . 3t y = (2C1t + C1 + C2)e . dx  = x − 2y − z  dt . . dy 183) Gia'i h^e. phuong trnh: = y − x + z  dt dz  = x − z dt 1 − λ −2 −1 . . . HD gia’i: Phuong trnhda. c trung −1 1 − λ 1 = 0 ⇔ λ(λ2 − λ − 2) = 0 1 0 −1 − λ ⇔ λ = 0, λ = −1, λ = 2 1 2 3     1 − λi −2 −1 P1i Vo.i cac gia'i h^e: λi; i = 1, 2, 3 .  −1 1 − λi 1  P2i = 0 1 0 −1 − λ P . . . . i 3i. D- ^e' tmnghi^e.m ri^engtuong ung. Tu dosuy ra h^e. nghi^e.m co ba'n: −t −t 2t x1 = 1, y1 = 0, z1 = 1; x2 = 0, y2 = e , z2 = −2e ; x3 = 3e , y = −2e−2t, z = e2t. 3 3  x = C + 3C e2t  1 3 ' −t 2t V^a. y h^e. nghi^e.m t^ong quat: y = C2e − 2C3e  −t 2t z = C1 − 2C2e + C3e dx  − 5x − 3x = 0 Gia'i h^e phu.o.ng trnh:  dt 184) . dy  + 3x + y = 0  dt . . . 5 − λ 3 HD gia’i: Phuong trnh da. c trung = 0 ⇔ λ = 2 (b^o. i 2) −3 −λ − 1
  42. 20 www.VNMATH.com  a − 3b = 3d at + b  ⇒ nghi^e.m coda.ng e2t thay vaoh^e. ⇒ a + c = 0 ct + d c + 3b = −3d C Cho a = C , b = C ⇒ c = −C , d = 1 − C 1 2 1 3 2  2t x = (C1t + C2)e V^a. y nghi^e.m t^o'ng quat: C y = (−C t + 1 − C )e2t.  1 3 2 dx  = 2x − 3y Gia'i h^e phu.o.ng trnh:  dt 185) . dy  = x − 2y + 2 sin t  dt . . . HD gia’i: Phuong trnh da. c trung cohai nghi^e.m λ1,2 = ±1       ' 0 3 −3 γ11 + λ1 = −1 giai h^e.: = ⇒ γ11 = γ12 = 1. 0 1 −1 γ12 1 −3 γ  0 + λ = 1 gia'i h^e.: 21 = ⇒ γ = 3; γ = 1. 2 1 −3 γ 0 21 22 . 22 . . . H^e. nghi^e.m co ba'n cu'a h^e. thu^an nh^at tuong ung la: ( −t ( t x1 = e x2 = 3e −t ; t y1 = e y2 = e ( −t t V^ay NTQ cu'a h^e thu^an nh^at: x(t) = C1e + 3C2e . . −t t y(t) = C1e + C2e Bi^en thi^enhang s^o:  ( ( 3 t C0 e−t + 3C0 et = 0 C0 = 3et sin t C1(t) = e (sin t − cos t) 1 2 ⇒ 1 ⇒ 2 0 −t 0 t 0 −t 1 C1e + C2e = 2 sin t C2 = e sin t C (t) = − e−t(sin t + cos t)  2 2 ( −t t V^ay NTQ: x(t) = C1e + 3C2e − 3 cos t . −t t y(t) = C1e + C2e + sin t − 2 cos t dx  = 2x − y + z  dt . . dy 186) Gia'i h^e. phuong trnh: = x + 2y − z  dt dz  = x − y + 2z dt . . . HD gia’i: Phuong trnh da. c trung: (λ − 1)(λ − 2)(λ − 3) = 0 co3 nghi^e.m λ1 = 1; λ2 = 2; λ3 = 3.       2 − λi −1 1 P1i 0 u.ng vo.i gia'i h^e: λi .  1 2 − λi −1  P2i = 0 1 −1 2 − λi P3i 0 0 1 1  0  e2t e3t - . . . t 2t Duo. c 1 ; 1 ; 0. Suy ra h^e. nghi^e.m co ba'n e  ; e  ;  0  1 1 1 et e2t e3t
  43. www.VNMATH.com 21  x = C e2t + C e3t  2 3 t 2t V^a. y NTQ: y = C1e + C2e  t 2t 3t z = C1e + C2e + C3e . dx  = y − 5 cos t Gia'i h^e phu.o.ng trnh:  dt 187) . dy  = 2x + y  dt . . . . . . HD gia’i: Dungphuong phap khu' : L^ay da.o hamtheo t phuong trnh thu hai: y” = 2x0 + y0 . . . D- ^e' yphuong trnh d^au, dua v^e: y” = 2(y − 5 cos t) + y0 ⇔ y” − y0 − 2y = −10 cos t. . . . . D- ^aylaphuong trnh tuy^en tnh c^ap hai, gia'i ra duo. c nghi^e.m t^o'ng quat: 2t −t y = C1e + C2e + 3 cos t + sin t . . 1 Thay vaophuong trnhd^au: x = C e2t − C e−t − cos t − 2 sin t 2 1 2 ( 2t −t V^ay NTQ: x = A1e + A2e − cos t − 2 sin t . 2t −t y = 2A1e − A2e + 3 cos t + sin t. ( . . y0 = 3y + 2z + 4e5x 188) Gia'i h^e. phuong trnh: z0 = y + 2z ( x 4x . . . y = C1e + 2C2e ’ Nghi^em phuong trnhdac trung ; NTQ: HD giai: . . λ1 = 1; λ2 = 4 x 4x z = −C1e + C2e  4 0 4x ( x 4x 5x C1 = e y = C e + 2C e + 3e Bi^en thi^enhang s^o: 3 → NTQ 1 2 4 x 4x 5x C0 = ex z = −C1e + C2e + e  2 3 ( . . y0 = 2y − z + 2ex 189) Gia'i h^e. phuong trnh: z0 = 3y − 2z + 4ex ( y = C ex + C e−x ’ Nghi^em t^o'ng quat cu'a h^e thu^an nh^at: 1 2 HD giai: . . x −x z = C1e + 3C2e ( y∗ = xex Nghi^e.m ri^engcu'a h^e. kh^ongthu^an nh^at: z∗ = (x + 1)ex. ( x −x x V^ay nghi^em t^o'ng quat: y = C1e + C2e + xe . . x −x x z = C1e + 3C2e + (x + 1)e . ( . . y0 = 2y − 4z + 4e−2x 190) Gia'i h^e. phuong trnh: z0 = 2y − 2z
  44. 22 www.VNMATH.com ( y = C (cos 2x − sin 2x) + C (cos 2x + sin 2x) ’ Nghi^em t^o'ng quat: 1 2 HD giai: . −2x z = C1 cos 2x + C2 sin 2x + e . dy  = y + z Gia'i h^e phu.o.ng trnh:dx 191) . dz  = z − 4y. dx . . . 2 HD gia’i: Phuong trnhda. c trung λ − 2λ + 5 = 0. Khi do λ1 = 1 + 2i, λ2 = 1 − 2i. H^e. thu^an nh^at conghi^e.m x x y = e (C1 cos 2x + C2 sin 2x), z = 2e (C2 cos 2x − C1 sin 2x). dy  = y + z + ex Gia'i h^e phu.o.ng trnh:dx 192) . dz  = z − 4y. dx . . . 2 HD gia’i: Phuong trnhda. c trung λ − 2λ + 5 = 0. Khi do λ1 = 1 + 2i, λ2 = 1 − 2i. H^e. thu^an nh^at conghi^e.m x x y = e (C1 cos 2x + C2 sin 2x), z = 2e (C2 cos 2x − C1 sin 2x). Vanghi^e.m cu'a h^e. kh^ongthu^an nh^at x x x y = e (C1 cos 2x + C2 sin 2x), z = 2e (C2 cos 2x − C1 sin 2x) − e . dy  = 2y − z Gia'i h^e phu.o.ng trnh:dx 193) . dz  = 2z + 4y + e2x. dx . . . 2 HD gia’i: Phuong trnhda. c trung λ − 4λ + 8 = 0. Khi do λ1 = 2 + 2i, λ2 = 2 − 2i. H^e. thu^an nh^at conghi^e.m 2x 2x y = e (C1 cos 2x + C2 sin 2x), z = −2e (C2 cos 2x − C1 sin 2x). Vanghi^e.m cu'a h^e. kh^ongthu^an nh^at 1 y = e2x(C cos 2x + C sin 2x) − e2x, z = −2e2x(C cos 2x − C sin 2x). 1 2 4 2 1 dy  = 2y + z + ex Gia'i h^e phu.o.ng trnh:dx 194) . dz  = z − 4y. dx HD gia’i: . . . 2 Phuong trnh da. c trung λ − 2λ + 5 = 0. Khi do λ1 = 1 + 2i,
  45. www.VNMATH.com 23   λ2 = 1 − 2i. H^e. thu^an nh^at conghi^e.m x x y = e (C1 cos 2x + C2 sin 2x), z = 2e (C2 cos 2x − C1 sin 2x). Vanghi^e.m cu'a h^e. kh^ongthu^an nh^at x x x y = e (C1 cos 2x + C2 sin 2x), z = 2e (C2 cos 2x − C1 sin 2x) − e . dx  = x + 2y Gia'i h^e phu.o.ng trnh:  dt 195) . dy  = x − 5 sin t.  dt . . HD gia’i: Nghi^e.m t^o'ng quat cu'a h^e. phuong trnh thu^an nh^at: ( −t 2t x = C1e + 2C2e t 2t y = −C1e + C2e . . . Bi^en thi^enhang s^o d^e' duo. c nghi^e.m: ( −t 2t 8 4 x = C1e + 2C2e + 3 sin t + 3 cos t t 2t y = −C1e + C2e + 2 cos t − sin t. dx  = x − 2y + et Gia'i h^e phu.o.ng trnh:  dt 196) . dy  = x + 4y + e2t.  dt ' . . . . . . ' HD gia’i: Nghi^e.m cua phuong trnhda. c trung: r1 = 2; r2 = 3; tu doduo. c NTQ cua ( 2t 3t h^e phu.o.ng trnh thu^an nh^at la: x = 2C1e + C2e . 2t 3t y = −C1e − C2e . . . Bi^en thi^enhang s^o d^e' duo. c nghi^e.m t^o'ng quat cu'a h^e. kh^ongthu^an nh^at:  3 x = 2C e2t + C e3t − et + 2te2t  1 2 2 1 y = −C e2t − C e3t + et − (t + 1)e2t.  1 2 2 dx  = 2x + y Gia'i h^e phu.o.ng trnh:  dt 197) . dy  = 4y − z.  dt ' . . . HD gia’i: Nghi^e.m cua phuong trnhda. c trung: r1 = r2 = 3. V^a. y NTQ coda.ng: ( 3t x = (λ1 + µ1t)e vo.i 3t λ2 = λ1 + µ1; µ2 = µ1 y = (λ2 + µ2t)e . ( 3t Tu.c la: x = (C1 + C2t)e 3t y = (C1 + C2 + C2t)e .
  46. 24 www.VNMATH.com dx  = 3x + 8y Tmnghi^em cu'a h^e phu.o.ng trnh:  dt 198) . . dy  = −x − 3y  dt tho'a ma~n cac di^eu ki^e.n: x(0) = 6; y(0) = −2 . . . . dy HD gia’i: Tu phuong trnhthu hai: x = − − 3y, l^ay da.o hamtheo t hai v^e, r^oi dt . . . . . d2y thay vaophuong trnhthu nh^at cu'a h^e. duo. c: − y = 0, gia'i ra: y = C et − C e−t, dt 1 2 t −t suy ra x = −4C1e − 2C2e ' ~  ' thoa man cac di^eu ki^e.n x(0) = 6; y(0) = −2, suy ra C1 = C2 = −1. V^a. y nghi^e.m cua h^e.: ( x = 4et + 2e−t y = −et − e−t dx  = 3x − y + z  dt . . dy 199) Gia'i h^e. phuong trnh: = −x + 5y − z  dt dz  = x − y + 3z. dt . . . 3 2 ' HD gia’i: Phuong trnhda. c trung: λ − 11λ + 36λ − 36 = 0, giai ra λ1 = 2; λ2 = . . . . ' 3; λ3 = 6. Tu doduo. c ba h^e. nghi^e.m co ban:  e2t   0   −e2t   e3t  ;  e3t  ;  e3t  . e6t −2e6t e6t V^a. y nghi^e.m t^o'ng quat:  x = C e2t + C e3t + C e6t  1 2 3 3t 6t y = C2e − 2C3e  2t 2t 6t z = −C1e + C2e + C3e . dy  = y + z Tmnghi^em t^o'ng quat cu'a h^e phu.o.ng trnh: dx 200) . . dz  = z − 4y. dx . . . . HD gia’i: Phuong trnhda. c trung: (λ − 1)(λ − 2) = 0, gia'i ra λ = 1; λ = 2. Tu do . . . 1 2 duo. c ba h^e. nghi^e.m co ba'n:  ex   2e2x  ; . −ex −3e2x V^a. y nghi^e.m t^o'ng quat: ( x 3x y = C1e + 2C2e x 2x z = −C1e − 3C2e .
  47. www.VNMATH.com 25 dx  = 2x − 3y Gia'i h^e phu.o.ng trnh:  dt 201) . dy  = x − 2y + 2 sin t.  dt . . . . . . HD gia’i: Phuong trnhda. c trung cocac nghi^e.m λ1 = −1; λ2 = 1. Tu doduo. c h^e. .  e−t   3et  nghi^e.m co ba'n: ; . e−t et ( −t t V^ay nghi^em t^o'ng quat: x = C1e + 3C2e . . −t t y = C1e + C2e . ( 0 −t 0 t ( 0 t Bi^en thi^enhang s^o: C 1e + 3C 2e = 0 C 1 = 3e sin t 0 −t 0 t ⇐⇒ 0 −t C 1e + C 2e = 2 sin t. C 2 = e sin t.  3 C (t) = et(sin t − cos t) Gia'i ra:  1 2 1 C (t) = − e−t(sin t + cos t).  2 2 ( −t t V^ay nghi^em t^o'ng quat cu'a h^e: x(t) = C1e + 3C2e − 3 cos t . . . −t t y(t) = C1e + C2e + sin t − 2 cos t.