Phương trình vi phân có lời giải chi tiết

thu^o.c tuy^en tnh.. Tnh di.nh thu.c Wronski cu'a chung... d^o.c l^a.p tuy^en tnh.Tnh di.nh thu.c Wronski cu'a chung.

2) Gia'i phu.o.ng trnh: √y.y” = y0

HD gia’i: D- a.t y0 = p ⇒ y” = pdp

dy (ham theo y) Phu.o.ng trnh tro.' thanh: √ypdp

dy = pVo.i p 6= 0 ta du.o c phu.o.ng trnh: dp = √dy

3) Gia'i phu.o.ng trnh: a(xy0+ 2y) = xyy0

HD gia’i: a(xy0+ 2y) = xyy0 ⇒ x(a − y)y 0 = −2ay

N^eu y 6= 0, ta co phu.o.ng trnh tu.o.ng du.o.ng vo.i a − y

y dy = −

2a

x dx ⇔ x

2a y a e−y = CNgoai ra y = 0 cu~ng la nghi^e.m

4) Gia'i phu.o.ng trnh: y” = y0ey

HD gia’i: D- a.t y0 = p ⇒ y” = pdp

dy thay vao phu.o.ng trnh: pdp

Ngoai ra y = C : hang la m^o.t nghi^e.m

5) Gia'i phu.o.ng trnh: xy0 = y(1 + ln y − ln x) vo.i y(1) = e

HD gia’i: D- u.a phu.o.ng trnh v^e: y0 =

6) Gia'i phu.o.ng trnh: y”(1 + y) = y02+ y0

HD gia’i: D- a.t y0 = z(y) ⇒ z0 = zdz

dy thay vao phu.o.ng trnh: dz

Tom la.i nghi^e.m t^o'ng quat: y = C, y = C − x; 1

C1 ln |C1y + C1− 1| = x + C2

7) Gia'i phu.o.ng trnh: y0 = y 2 − 2

x 2

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:

8) Gia'i phu.o.ng trnh: yy” + y02 = 1

HD gia’i: D- a.t y0 = z(y) ⇒ y” = z.dz

dyBi^en d^o'i phu.o.ng trnh v^e: z

Nghi^e.m t^o'ng quat: y 2 + C1 = (x + C2) 2

9) Gia'i phu.o.ng trnh: 2x(1 + x)y0 − (3x + 4)y + 2x√1 + x = 0

Cx2

√

x + 1

Bi^en thi^en hang s^o: C0 = − 1

y0(0) = 0

HD gia’i: D- a.t z = y0 → y” = z.dz

dy phu.o.ng trnh tro.' thanh z.dz

√

e 2y − 1 = x + ε d¯ˆo’i biˆe´n t =

√

e 2y − 1 arctg √

e 2y − 1 = x + ε y(0) = 0 ⇒ ε = 0. V^a.y nghi^e.m ri^eng thoa' di^eu ki^e.n d^e bai: y = 1

2ln(tg

2 x + 1).

11) Tm nghi^e.m ri^eng cu'a phu.o.ng trnh: xy0+ 2y = xyy0

thoa' ma~n di^eu ki^e.n d^au y(−1) = 1

HD gia’i: Vi^et phu.o.ng trnh la.i: x(1 − y)y0 = −2y; do y(−1) = 1 n^en y 6≡ 0 D- u.a v^ephu.o.ng trnh tach bi^en: 1 − y

y dy = −2

dx xtch ph^an t^o'ng quat: x 2 ye−y = C Thay di^eu ki^e.n vao ta du.o c C = 1

e V^a.y tch ph^anri^eng c^an tm la: x 2 ye 1−y = 1

12) Bang cach da.t y = ux, ha~y gia'i phu.o.ng trnh: xdy − ydx − px 2 − y 2 dx = 0 (x > 0)

HD gia’i: D- a.t y = ux; du = udx + xdu thay vao phu.o.ng trnh va gia'n u.o.c x: xdu −

13) Tm nghi^e.m ri^eng cu'a phu.o.ng trnh: xy0 = px 2 − y 2 + y

thoa' ma~n di^eu ki^e.n d^au y(1) = 0

⇐⇒ 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) Tm nghi^e.m ri^eng cu'a phu.o.ng trnh: y0sin x = y ln y

thoa' ma~n di^eu ki^e.n d^au y(π

⇐⇒ ln y = C tanx

2 ⇐⇒ y = eC tan

x 2

thoa' ma~n di^eu ki^e.n d^au y(π

2) = e khi C = 1 V^a.y y = etan

x

2

15) Tm nghi^e.m ri^eng cu'a phu.o.ng 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

phu.o.ng trnh thanh: (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

16) Bang cach da.t y = 1

z r^oi da.t z = ux,ha~y gia'iphu.o.ng trnh: (x 2 y 2 − 1)dy + 2xy 3 dx = 0

18) Tm nghi^e.m t^o'ng quat cu'a cac phu.o.ng trnh sau: y0− y = y 2

HD gia’i: D- ^ay la phu.o.ng trnh tach bi^en va co nghi^e.m t^o'ng quat la

20) Tm nghi^e.m cu'a cac phu.o.ng trnh sau: y0− y = y 3

HD gia’i: D- ^ay la phu.o.ng trnh tach bi^en va co nghi^e.m t^o'ng quat la

x + C ⇔ sin z = − ln |x| + CV^a.y TPTQ: siny

x = − ln |x| + C

23) Gia'i phu.o.ng trnh: (y02− 1)x 2 y 2 + y0(x 4 − y 4 ) = 0

HD gia’i: La phu.o.ng trnh da'ng c^ap nhu.ng gia'i kha phu.c ta.p

Xem phu.o.ng trnh b^a.c hai d^oi vo.i y0: 4 = (x 4 + y4)2 ⇒ y 0

1 = y

x 2 ; y20 = −x

y 2 Tu do co hai ho nghi^e.m t^o'ng quat: y = x

C1x + 1; x

3 + y 3 = C2

24) Gia'i phu.o.ng trnh: y2+ x2y0 = xyy0

HD gia’i: Vi^et phu.o.ng trnh la.i y0 =

y 2

x 2

y

x − 1 d^ay la phu.o.ng trnh thu^an nh^at, gia'i

ra du.o c nghi^e.m t^o'ng quat: y2 = Cxeyx

25) Tm nghi^e.m ri^eng cu'a phu.o.ng trnh: (x + y − 2)dx + (x − y + 4)dy = 0

thoa' ma~n di^eu ki^e.n d^au y(1) = 0

HD gia’i: D- a.t

(

x = u − 1

y = v + 3. thay vao phu.o.ng trnh du.o c:

(u + v)du + (u − v)dv = 0, d^ay la phu.o.ng trnh thu^an nh^at co tch ph^an t^o'ng quat la:

u 2 + 2uv − v 2 = C

V^a.y tch ph^an t^o'ng quat cu'a phu.o.ng trnh ban d^au la: x 2 + 2xy − y 2 − 4x + 8y = C

26) Gia'i phu.o.ng trnh (x + y − 2)dx + (x − y + 4)dy = 0

X +

1 − u

1 + 2u − u 2 du = 0.Gia'i ra X 2 (1 + 2u − u 2 ) = C hay x 2 + 2xy − y 2 − 4x + 8y = C

27) Tm tch ph^an t^o'ng quat cu'a phu.o.ng trnh sau: b) y0 = 2xy

x 2 − y 2

HD gia’i: D- ^ay la phu.o.ng trnh da'ng c^ap, ta da.t z = y

z Khi do phu.o.ng trnh tr^entro.' thanh xz0 = z(1 + z

V^a.y nghi^e.m cu'a phu.o.ng trnh da~ cho la x 2 + y 2 = C1y, C1 6= 0.

28) Tm nghi^e.m t^o'ng quat cu'a cac phu.o.ng trnh sau: y0 = 2x + y − 1

4x + 2y + 5.

HD gia’i: D- a.t u = 2x + y phu.o.ng trnh du.a v^e da.ng

du

dx =5u + 9 2u + 5.

Gia'i phu.o.ng trnh nay ta du.o c nghi^e.m 10u + 7 ln |5u + 9| = 25x + C.

V^a.y nghi^e.m cu'a phu.o.ng trnh da~ cho la 10y + 7 ln |10x + 5y = 9| − 5x = C.

29) Tm tch ph^an t^o'ng quat cu'a cac phu.o.ng trnh sau:

V^a.y nghi^e.m cu'a phu.o.ng trnh da~ cho la y 2 − x 2 − 2xy − 8y + 4x = C1.

30) a) Tm mi^en ma trong do nghi^e.m cu'a bai toan Cauchy cu'a phu.o.ng trnh

sau d^ay t^on ta.i va duy nh^at y0 = √

x − y.

b) Tm tch ph^an t^o'ng quat cu'a cac phu.o.ng trnh sau: (x 2 − y 2 )dy − 2xydx = 0.

HD gia’i:

a) Bai toan Cauchy co duy nh^at nghi^e.m trong mi^en

D = {(x, y) ∈ R 2 |x − y ≥ δ} vo.i δ > 0 tuy y

b) D- u.a phu.o.ng trnh v^e da.ng dy

z − 2z

1 + z 2 )dz = dx

x Suy ra nghi^e.m cu'a phu.o.ng trnh nay la 1 + zz 2 = Cx, C 6= 0.

V^a.y nghi^e.m cu'a phu.o.ng trnh da~ cho la x 2 + y 2 = C1y, C1 6= 0.

31) a) Chu.ng minh rang h^e cac vecto.{e 2x , xe2x, x2} la h^e d^o.c l^a.p tuy^en tnh

b) Tm tch ph^an t^o'ng quat cu'a phu.o.ng trnh sau: (x − y)dy − (x + y)dx = 0;

HD gia’i:

a) Dung di.nh ngh~a ki^e'm tra h^e d^o.c l^a.p tuy^en tnh

b) D- u.a phu.o.ng trnh v^e da.ng y0 = x + y

x − y D- ^ay la phu.o.ng trnh da'ng c^ap, ta da.t

p

x 2 + y 2 = Cearctgyx

32) a) Chu.ng minh rang h^e cac vecto.{cos 2 2x, sin22x, 2} la h^e phu thu^o.c tuy^en tnh

Tnh di.nh thu.c Wronski cu'a chung

b) Tm tch ph^an t^o'ng quat cu'a phu.o.ng trnh sau: (x − 2y + 1)dy − (x + y)dx = 0.

HD gia’i:

a) H^e nay phu thu^o.c tuy^en tnh v 2 cos 2 2x + 2 sin22x − 2 = 0

b) Phu.o.ng trnh nay co th^e' du.a v^e da.ng da'ng c^ap, ta du.o c

.Hay p(3x − 1) 2 + 2(3y + 1) 2 = C1e√12 arctg(√23x−13y+1)

.

33) Gia'i phu.o.ng trnh: y 2 + x 2 y0 = xyy0

HD gia’i: Phu.o.ng trnh thu^an nh^at: da.t y = zx → y0 = z0x + z

Phu.o.ng trnh tro.' thanh z − 1

34) Gia'i phu.o.ng trnh y 2 + x 2 y0 = xyy0

HD gia’i: Vi^et phu.o.ng trnh la.i y0 =

y 2

x 2

y

x − 1 d^ay la phu.o.ng trnh thu^an nh^at, gia'i

ra du.o c nghi^e.m t^o'ng quat: y 2 = Cxeyx

35) Gia'i phu.o.ng trnh: y” cos y + (y0) 2 sin y = y0

HD gia’i: y = C : hang la m^o.t nghi^e.m

y 6= C (hang) D- a.t y0 = p ⇒ y” = pdp

dy (ham theo y)thay vao (2): dp

dycos y + p sin y = 1: phu.o.ng trnh tuy^en tnh

Phu.o.ng trnh thu^an nh^at co nghi^e.m t^o'ng quat: p = C cos y.

bi^en thi^en hang s^o du.o c C = tgy + C1

tu do p = dy

dx = sin y + C1cos y ⇔

dy sin y + C1cos y = dx

tch ph^an di d^en: 1

pC 2

1 + 1ln

+ 1

C1

= x + C2

36) Gia'i phu.o.ng trnh: y0+ 1

2x − y 2 = 0

HD gia’i: Coi x = x(y) la ham cu'a y ta co: y0 = 1

x 0 thay vao phu.o.ng trnh:

x 0 + 1

2x − y 2 = 0 ⇔ x0+ 2x = y 2 : phu.o.ng trnh tuy^en tnh

Nghi^e.m t^o'ng quat cu'a phu.o.ng trnh thu^an nh^at: x = Ce−2y

Bi^en thi^en hang s^o: C0(y) = y 2 e 2y ⇒ C(y) = 1

2y

2 − 1

2y +

1 4

37) Gia'i phu.o.ng trnh: xy” = y0+ x 2

HD gia’i: D- a.t y0 = p, (1) tro.' thanh: xp0 − p = x 2 tuy^en tnh

Nghi^e.m t^o'ng quat cu'a phu.o.ng trnh thu^an nh^at: p = Cx

Bi^en thi^en hang s^o → C(x) = x + C1

38) Gia'i phu.o.ng trnh: y02+ yy” = yy0

HD gia’i: D- a.t p = y0(p 6= 0), phu.o.ng trnh tu.o.ng du.o.ng vo.i: p 2 + ypdp

y, bi^en thi^en hang s^o

⇒ C(y) = y

2

2 + C1Nhu v^a.y: p = y

2 + 2C12y ⇒ dy

dx =

y 2 + 2C12y ⇒ 2ydy

y 2 + 2C 1

= dx

⇒ y 2 = A1e x + A2.

Chu y: V^e trai (yy0)0 = yy0 ⇔ yy 0 = C1e x ⇔ ydy = C1e x dx ⇔ y 2 = 2C1e x + C2

39) Gia'i phu.o.ng trnh: ye y = y0(y 3 + 2xe y ) vo.i y(0) = −1

HD gia’i: yx0 = 1

x 0 y bi^en d^o'i phu.o.ng trnh v^e: x0− 2

yx = y

2 e−yNghi^e.m t^o'ng quat: x = y2(C − e−y)

y(0) = −1 ⇒ C = e.

V^a.y x = y 2 (e − e−y)

40) Gia'i phu.o.ng trnh: xy” = y0+ x

HD gia’i: D- a.t y0 = p; phu.o.ng trnh tro.' thanh: p0 − 1

xp = 1Nghi^e.m t^o'ng quat: p = Cx bi^en thi^en hang s^o: C = ln |x| + C1

⇒ p = dy

dx = (ln |x| + C1)x ⇒ y =

Z (ln |x| + C1)xdx + C2

HD gia’i: Nghi^e.n t^o'ng quat cu'a phu.o.ng trnh thu^an nh^at y = Ce−x22

bi^en thi^en hang s^o: C(x) = (x 2 − 2)e − x2

2 + εV^a.y nghi^e.m t^o'ng quat: y = εe−x22 + x 2 − 2.

42) Gia'i phu.o.ng trnh: (x 2 − y)dx + xdy = 0

HD gia’i: Phu.o.ng trnh vi^et la.i: xy0− y = −x 2, phu.o.ng trnh thu^an nh^at: xy0− y = 0co nghi^e.m t^o'ng quat: y = Cx bi^en thi^en hang s^o suy ra C = −x + ε

V^a.y nghi^e.m t^o'ng quat : y = −x 2 + εx

x

44) Gia'i phu.o.ng trnh: (x + 1)(y0+ y 2 ) = −y

HD gia’i: Xet y 6= 0, bi^en d^o'i phu.o.ng trnh v^e da.ng y0+ 1

ngoai ra y = 0 cu~ng la nghi^e.m

V^a.y nghi^e.m t^o'ng quat: y = 1

(x + 1)(ln |x + 1| + ε) va y = 0 nghi^e.m k di

45) Gia'i phu.o.ng trnh: 2xy0+ y = 1

1 − x

HD gia’i: D- u.a phu.o.ng trnh v^e da.ng y0 + 1

2xy =

1 2x(1 − x) phu.o.ng trnh tuy^entnh c^ap 1

Nghi^e.m t^o'ng quat: y = √C

x, bi^en thi^en hang s^o:

C0(x) =

√ x 2x(1 − x) ⇒ C = 1

Nghi^e.m t^o'ng quat: y = (C − cos x)x

47) Gia'i phu.o.ng trnh: y0cos 2 x + y = tgx thoa' y(0) = 0

HD gia’i: Phu.o.ng trnh tuy^en tnh → NTQ y = Ce−tgx; y = tgx − 1 (m^o.t nghi^e.mri^eng)

⇒ NTQ: y = Ce−tgx+ tgx − 1

y(0) = 0 ⇒ C = 1 V^a.y nghi^e.m ri^eng c^an tm: y = tgx − 1 + e−tgx.

48) Gia'i phu.o.ng trnh: y0√

1 − x 2 + y = arcsin x thoa' y(0) = 0

HD gia’i: Nghi^e.m t^o'ng quat cu'a phu.o.ng trnh tuy^en tnh thu^an nh^at: y = Ce−arcsinxD^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^eng c^an tm: y = e−arcsinx+ arcsinx − 1

49) Tm nghi^e.m ri^eng cu'a phu.o.ng trnh: y0 = 1

2x − y 2

thoa' ma~n di^eu ki^e.n d^au y(1) = 0

HD gia’i: Xem x la ^a'n ham, thay y0 = 1

x 0, phu.o.ng trnh thanh1

x 0 = 12x − y 2 ⇐⇒ x0− 2x = −y 2

D- ^ay la phu.o.ng trnh tuy^en tnh c^ap m^o.t, nghi^e.m t^o'ng quat cu'a phu.o.ng trnh tuy^entnh thu^an nh^at tu.o.ng u.ng la x = Ce−2y Bi^en thi^en hang s^o du.o c NTQ:

4.V^a.y nghi^e.m tho'a ma~n di^eu ki^e.n d^au: x = 3

4

50) Gia'i phu.o.ng trnh sau d^ay, bi^et rang sau khi da.t y = z

x 2, ta nh^a.n du.o cm^o.t phu.o.ng trnh vi ph^an c^ap hai co m^o.t nghi^e.m ri^eng y∗ = 1

x

2 , NTQ cu'a phu.o.ng trnh thu^an nh^at:

z = C1cos x + C2sin x V^a.y NTQ cu'a phu.o.ng trnh ban d^au la:

51) Tm nghi^e.m ri^eng cu'a phu.o.ng trnh: ye y = y0(y 3 + 2xe y )

thoa' ma~n di^eu ki^e.n d^au y(0) = −1

HD gia’i: Xem x la ^a'n ham, thay y0 = 1

x 0, phu.o.ng trnh thanh x0 − 2

yx = y

2 e−y.NTQ cu'a phu.o.ng trnh tuy^en tnh thu^an nh^at tu.o.ng u.ng la x = C

y; bi^en thi^en hangs^o du.o c C(y) = −e−y+ C Nhu v^a.y NTQ la x = C

y − 1

ye y Thay di^eu ki^e.n d^au xac di.nhdu.o c C = 1

e Tu do KL

52) Tm nghi^e.m cu'a phu.o.ng 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 phu.o.ng trnh tuy^en tnh ra y = Ce x + sin x

tho'a di^eu ki^e.n y bi cha.n khi x → ∞ khi C = 0

53) Tm nghi^e.m ri^eng cu'a phu.o.ng trnh: y0+ sin y + x cos y + x = 0

thoa' ma~n di^eu ki^e.n d^au y(0) = π

, phu.o.ng trnh thanh phu.o.ng trnh tuy^en tnh

z0 + z = −x Gia'i ra: z = 1 − x + Ce−x

thoa' ma~n di^eu ki^e.n d^au y(0) = π

2 khi C = 0 V^a.y nghi^e.m ri^eng y = 2 arctan(1 − x)

54) Tm nghi^e.m t^o'ng quat cu'a cac phu.o.ng trnh sau: y0− x tan y = x

cos y

HD gia’i: D- a.t z = sin y, khi do phu.o.ng trnh da~ cho tro.' thanh z0− xz = x. D- ^ay laphu.o.ng trnh tuy^en tnh c^ap 1 va co nghi^e.m t^o'ng quat la z = Cex22 − 1

V^a.y nghi^e.m cu'a phu.o.ng trnh da~ cho la sin y = z = Cex22 − 1

55) Tm nghi^e.m t^o'ng quat cu'a cac phu.o.ng trnh sau: y0− xy = x

HD gia’i: D- ^ay la phu.o.ng trnh Bernoulli va co nghi^e.m t^o'ng quat la

x − cos x.

60) Tm nghi^e.m cu'a cac phu.o.ng trnh sau: y0− y = x√y.

D- ^ay la phu.o.ng trnh vi ph^an tuy^en tnh c^ap 1

Nghi^e.m t^o'ng quat la y = (C + x

HD gia’i: D- ^ay la phu.o.ng trnh Bernoulli va co nghi^e.m la

√

y = 1

2ln x + Cx

2

63) a) Tm mi^en ma trong do nghi^e.m cu'a bai toan Cauchy cu'a phu.o.ng trnh sau

d^ay t^on ta.i va duy nh^at y0 = y + 3x.

b) Tm nghi^e.m cu'a bai toan Cauchy sau d^ay

D- ^ay la phu.o.ng trnh vi ph^an tuy^en tnh c^ap 1

Nghi^e.m t^o'ng quat la:

y = (C + x) cos x.

65) Tm nghi^e.m cu'a phu.o.ng trnh sau: y0+ y

66) Gia'i phu.o.ng trnh: (x + 1)y” + x(y0) 2 = y0

HD gia’i: D- a.t y0 = p, phu.o.ng trnh tro.' thanh phu.o.ng trnh Bernouili (vo.i x 6= −1)

x + 1Bi^en thi^en hang s^o cu^oi cung du.o c: z = x

2 + C12(x + 1) ⇒ y 0 = 1

z =

2(x + 1)

x 2 + C1Suy ra nghi^e.m t^o'ng quat cu'a phu.o.ng trnh:

67) Gia'i phu.o.ng trnh: x2y0 = y(x + y)

bi^en thi^en hang s^o C: C(x) = ε − 1

2x 2 V^a.y z = x(ε − 1

2x 2 )V^a.y nghi^e.m t^o'ng quat la: y = 2x

HD gia’i: D- a.t y = p(y); y = p.py thay vao phu.o.ng trnh

pydp

dy − p 2 = y3,da.t ti^ep: p(y) = y.z(y) du.a phu.o.ng trnh v^e

= x + C2.

do y(0) = −1

2 ⇒ C 2 = 0.

V^a.y nghi^e.m ri^eng c^an tm thoa' : ln

... gia’i: Phu.o.ng trnh vi ph^an toan ph^an: Nghi^e.m t^o''ng quat: x (1 + ln y) − y = C

88) Tm nghi^e.m t^o''ng quat cu''a phu.o.ng trnh vi ph^an: 3x (1... quat cu''a phu.o.ng trnh vi ph^an:

(sin xy + xy cos xy)dx + x2cos xydy = 0

HD gia’i: Phu.o.ng trnh vi ph^an toan ph^an co... y0− y = x√y.

D- ^ay la phu.o.ng trnh vi ph^an tuy^en tnh c^ap

Nghi^e.m t^o''ng quat la y = (C + x