Tổng hợp 201 Bài tập phương trình vi phân | Phương trình vi phân | Trường Đại học Khoa học Tự nhiên, Đại học Quốc gia Hà Nội
Tài liệu được sưu tầm và soạn thảo dưới dạng file PDF với mục đích hỗ trợ học tập và tham khảo. Nội dung tài liệu được trình bày rõ ràng, dễ tiếp cận, phù hợp cho việc ôn tập và củng cố kiến thức trong quá trình học đại học. Đây sẽ là nguồn tư liệu hữu ích giúp các bạn sinh viên chuẩn bị tốt hơn cho các buổi học, đồng thời mở rộng thêm hiểu biết về môn học. Hy vọng tài liệu này sẽ mang lại nhiều giá trị và hỗ trợ các bạn trong hành trình học tập. Mời bạn đọc cùng tham khảo!
Môn: Phương trình vi phân (HUS) 10 tài liệu
Trường: Trường Đại học Khoa học tự nhiên, Đại học Quốc gia Hà Nội 687 tài liệu
Tác giả:
Preview text:
1 www.VNMATH.com . . B ` AI T ˆ A . P PHU ONG TR`INH VI PH ˆ AN 1) 2xy′y” = y′2 − 1 HD gia’i: y′ = p : 2xpp′ = p2 − 1 2pdp dx √ x(p2 − 1) 6= 0 =
⇔ p2 − 1 = C1 ⇔ p = ± C1x + 1 p2 − 1 x dy √ 2 3 p = = C (C 2 + C dx 1 + 1 ⇒ y = 3C 1x + 1) 2 1 √ 2) y.y” = y′ dp √ dp HD gia’i: y′ = p ⇒ y” = p yp = p dy dy dy √ dy √ p 6= 0 dp = √ ⇒ p = 2 y + C = 2 y + C y 1 ⇔ dx 1 ⇒ dy dx = √ 2 y + C1 √ C √ x = y − 1 ln |2 y + C1| + C 2 2 y = c 3) a(xy′ + 2y) = xyy′
HD gia’i: a(xy′ + 2y) = xyy′ ⇒ x(a − y)y′ = −2ay a − y 2a y 6= 0 dy = − dx ⇔ x2ayae−y = C y x y = 0 4) y” = y′ey dp dp HD gia’i: y′ = p ⇒ y” = p p = pey dy dy dp dy dy p 6= 0 : = ey ⇔ p = ey + C = ey + C = dx dy 1 ⇒ dx 1 ⇔ ey + C1 dy 1 ey + C 1 eydy y C R 1 − ey 1 6= 0 R = dy = (y − R ) = − ey + C y 1 C1 ey + 1 C1 e + C1 C1 1 ln(ey + C1) C1 dx −e−y nˆe´u C 1 = 0 R = 1 ey + C1 (y − ln |ey + C1|) nˆe´u C1 6= 0. C1 y = C : 5) xy′ = y(1 + ln y − ln x) y(1) = e 2 www.VNMATH.com y y HD gia’i: y′ = (1 + ln ) y = zx xz′ = z ln z x x dz dx y • z ln z 6= 0 ⇒ = ⇒ ln z = Cx ln = Cx ⇔ y = xeCx z ln z x x y(1) = e → C = 1. y = xex 6) y”(1 + y) = y′2 + y′ dz dz dy HD gia’i: y′ = z(y) ⇒ z′ = z = dy z + 1 y + 1 dy
⇒ z + 1 = C1(y + 1) ⇒ z = C1y + C1 − 1 ⇔ = dx (∗) C1y + C1 − 1 • C1 = 0 ⇒ (∗) y = C − x 1 • C1 6= 0 ⇒ (∗) ln |C1y + C1 − 1| = x + C2 C1 y = C 1 y = C, y = C − x; ln |C1y + C1 − 1| = x + C2 C1 2 7) y′ = y2 − x2 HD gia’i: x2y′ = (xy)2 − 2 (∗) z = xy ⇒ z′ = y + xy′ (∗) dz dx r z − 1 xz′ = z2 + z − 2 ⇔ = ⇔ 3 = Cx z2 + z − 2 x z + x xy − 1 = Cx3. xy + 2 8) yy” + y′2 = 1 dz HD gia’i: y′ = z(y) ⇒ y” = z. dy z dy C dz = ⇔ z2 = 1 + 1 1 − z2 y y2 dy r C dy ⇒ = ± 1 + 1 ⇔ ± R = dx ⇒ y2 + C1 = (x + C dx y2 r 2)2 C 1 + 1 y2 y2 + C1 = (x + C2)2 √ 9)
2x(1 + x)y′ − (3x + 4)y + 2x 1 + x = 0 3x + 4 1 HD gia’i: y′ − .y = − √ ; x 6= 0, x 6= −1 2x(x + 1) x + 1 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 1 1 C′ = − ⇒ C = − + ε. x2 x x2 1 y = √ ( + ε) x + 1 x (y(0) = 0 10) y” = e2y y′(0) = 0 dz dz z2 e2y HD gia’i: z = y′ → y” = z. z. = e2y ⇔ = + ε dy dy 2 2 1
y′(0) = y(0) = 0 ⇒ ε = − . z2 = e2y − 1. 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. y = ln(tg2x + 1). 2 11) xy′ + 2y = xyy′ y(−1) = 1 HD gia ’ i: x(1 − y)y′ = −2y y(−1) = 1 y 6≡ 0 1 − y dx dy = −2 y x 1 x2ye−y = C C = e x2ye1−y = 1 12) y = ux
xdy − ydx − px2 − y2dx = 0. (x > 0) HD gia ’ i: y = ux; du = udx + xdu x xdu − √1 − u2dx = 0 u − ±1 u 6≡ ±1 du dx = arcsin u − ln x = C x > 0 1 − u2 x y y = ±x; arcsin = ln x + C x 13) xy′ = px2 − y2 + y y(1) = 0 HD gia’i: r y2 y
xy′ = px2 − y2 + y ⇐⇒ y′ = 1 − + x2 x y u = y = ux y′ = xu′ + u x √ du dx xu′ = 1 − u2 ⇐⇒ √ = 1 − u2 x 4 www.VNMATH.com ⇐⇒ arcsin u = ln Cx y(1) = 0 C = 1 y = ±x 14) y′ sin x = y ln y π y( ) = e 2 HD gia’i: dy dx y′ 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 y( ) = e C = 1 y = e 2 2 15)
(x + y + 1)dx + (2x + 2y − 1)dy = 0 y(0) = 1 HD gia’i: x + y = z =⇒ dy = dz − dx (2 − z)dx + (2z − 1)dz = 0
x − 2z − 3 ln |z − 2| = C
x + 2y + 3 ln |x + y − 2| = C y(0) = 1 C = 2 1 16) y = z = ux z (x2y2 − 1)dy + 2xy3dx = 0 1 HD gia ’ i: y = (z2 − x2)dz + 2zxdx = 0 z = ux 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 u = 1 + x2y2 = Cy xy 17) y′ − xy = x + x3 HD gia’i: 2 x2 x y = Ce 2 . + 1 2 5 www.VNMATH.com 18) y′ − y = y2. HD gia’i: y ln | | = x + C. y + 1 y 19) y′ + = ex x HD gia’i: C ex y = + ex − x x 20) y′ − y = y3. HD gia’i: C + x = ln |y| − arctgy. y y π 21) y′ = + sin y(1) = x x 2
HD gia’i: y = zx ⇒ y′ = z′x + z dz dx z z z′x = sin x ⇔ =
⇔ ln |tg | = ln |x| + ln C ⇔ tg = Cx sin z x 2 2 y π tg = Cx; y(1) = ⇒ C = 1. 2x 2 y tg = x 2x y y 22)
(x − y cos )dx + x cos dy = 0 x x y HD gia’i: = z ⇒ y′ = z′x + z x Z dx x cos z.z′ + 1 = 0 ⇔ cos zdz = − + C ⇔ sin z = − ln |x| + C x y sin = − ln |x| + C x 23)
(y′2 − 1)x2y2 + y′(x4 − y4) = 0 HD gia’i: 6 www.VNMATH.com y2 x2
y′ △ = (x4 + y4)2 ⇒ y′1 = ; y′ . x2 2 = − y2 x y = ; x3 + y3 = C C 2 1x + 1 24) y2 + x2y′ = xyy′ y2 HD gia ’ i: y′ = x2 y − 1 x y y2 = Cxex 25)
(x + y − 2)dx + (x − y + 4)dy = 0 y(1) = 0 (x = u − 1 HD gia’i: y = v + 3. (u + v)du + (u − v)dv = 0 u2 + 2uv − v2 = C
x2 + 2xy − y2 − 4x + 8y = C 26)
(x + y − 2)dx + (x − y + 4)dy = 0 (x = X − 1 HD gia’i: y = Y + 3 (X + Y )dX + (X − Y )dY = 0 dX 1 − u Y = uX + du = 0 X 1 + 2u − u2 X2(1 + 2u − u2) = C
x2 + 2xy − y2 − 4x + 8y = C 2xy 27) b) y′ = . x2 − y2 y HD gia ’ i: z = z z(1 + z2) 1 2z dx xz′ = ( − )dz = 1 − z2 z 1 + z2 x z = Cx, C 6= 0. 1 + z2 x2 + y2 = C1y, C1 6= 0. 2x + y − 1 28) y′ = . 4x + 2y + 5 HD gia’i: u = 2x + y du 5u + 9 = . dx 2u + 5 7 www.VNMATH.com 10u + 7 ln |5u + 9| = 25x + C.
10y + 7 ln |10x + 5y = 9| − 5x = C. 29)
(x − y + 4)dy + (y + x − 2)dx = 0 HD gia ’ i: x = dv u + v u + 1, y = v − 3, = du −u + v v2 − 2uv − v2 = C.
y2 − x2 − 2xy − 8y + 4x = C1. 30) √ y′ = x − y. (x2 − y2)dy − 2xydx = 0. HD gia’i:
D = {(x, y) ∈ R2|x − y ≥ δ} δ > 0 dy xy = dx x2 − y2 y z = x z(1 + z2) xz′ = . 1 − z2 1 2z dx ( − )dz = z 1 + z2 x z = Cx, C 6= 0. 1 + z2 x2 + y2 = C1y, C1 6= 0. 31) {e2x, xe2x, x2} (x − y)dy − (x + y)dx = 0; HD gia’i: x + y y′ = x − y y z = x 1 + z2 xz′ = . 1 − z px2 + y2 = Cearctgyx . 32) {cos2 2x, sin2 2x, 2}
(x − 2y + 1)dy − (x + y)dx = 0. 8 www.VNMATH.com HD gia’i:
2 cos2 2x + 2 sin2 2x − 2 = 0 x + y y′ = . x − 2y + 1 1 1 u = x − , v = y + 3 3 u + v v′ = . u − 2v √ √ 1 arctg( 2 u u2 + 2v2 = Ce √ ) 2 v . √ 1 p arctg( 2 3x−1 (3 )
x − 1)2 + 2(3y + 1)2 = C √2 3y+1 1e . 33) y2 + x2y′ = xyy′ HD gia’i: y = zx → y′ = z′x + z z − 1 dx dz = → z − ln |z| = ln |x| + C z x y y − ln | | = ln |x| + C x x 34) y2 + x2y′ = xyy′ y2 HD gia ’ i: y′ = x2 y − 1 x y y2 = Cxex 35)
y” cos y + (y′)2 sin y = y′ HD gia’i: y = C : dp y 6= C y′ = p ⇒ y” = p y dy dp cos y + p sin y = 1 dy p = C cos y. C = y C1 dy dy p = = sin y + C = dx dx 1 cos y ⇔ sin y + C1 cos y y r 1 1 tg + 1 + − 1 2 C2 C ln 1 1 = x + C p 2 C2 + 1 y r 1 1 1 −tg + 1 + + 2 C21 C1 1 36) y′ + = 0 2x − y2 1 HD gia’i: x = x(y) y y′ = x′ 9 www.VNMATH.com 1 1 + = 0 ⇔ x′ + 2x = y2 : x′ 2x − y2 x = Ce−2y 1 1 1
C′(y) = y2e2y ⇒ C(y) = y2e2y − ye2y + e2y + C 2 2 4 1 1 1 x = Ce−2y + y2 − y + 2 2 4 37) xy” = y′ + x2 HD gia’i: y′ = p xp′ − p = x2 p = Cx → C(x) = x + C1 dy x3 x2 = x(x + C + C + C2 dx 1) → y = 3 1. 2 38) y′2 + yy” = yy′ dp HD gia’i: p = y′(p 6= 0) p2 + yp = yp dy dp dp p ⇔ p + y = y y 6= 0 + = 1 dy dy y C p = y y2 ⇒ C(y) = + C 2 1 y2 + 2C dy y2 + 2C1 2ydy p = 1 ⇒ = ⇒ = dx 2y dx 2y y2 + 2C1 ⇒ y2 = A x 1e + A2.
(yy′)′ = yy′ ⇔ yy′ = C x x x 1e
⇔ ydy = C1e dx ⇔ y2 = 2C1e + C2 39) yey = y′(y3 + 2xey) y(0) = −1 1 2 HD gia’i: y′ = x′ x = y2e−y x − x′y y x = y2(C − e−y) y(0) = −1 ⇒ C = e. x = y2(e − e−y) 40) xy” = y′ + x 1 HD gia’i: y′ = p; p′ − p = 1 x p = Cx C = ln |x| + C1 10 www.VNMATH.com dy Z ⇒ p = = (ln |x| + C1)x ⇒ y = (ln |x| + C1)xdx + C dx 2 x2 x2 = C1x2 + ln |x| − + C 2 4 2 41) y′ + xy = x3 2 HD gia’i: y = Ce− x2 2 C(x) = (x2 − 2)e−x2 + ε 2 y = εe− x2 + x2 − 2. 42) (x2 − y)dx + xdy = 0 HD gia ’ i: xy′ −y = −x2 xy′ −y = 0 y = Cx C = −x + ε y = −x2 + εx 2 3 43) y′ − y = y(1) = 1 x x2 3 1 HD gia’i: y = Cx2; C′ = ⇒ C = − + ε x4 x3 1 y = εx2 − ; y(1) = 1 ⇒ ε = 2 x 1 y = 2x2 − x 44) (x + 1)(y′ + y2) = −y 1 HD gia’i: y 6= 0, y′ + .y = −y2 x + 1 1 z′ 1 = z ⇒ y′ = − = −y2z′ z′ − .z = 1. y z2 x + 1 z = C1(x + 1) C1 = ln |x + 1| + ε. z = (x + 1)(ln |x + 1| + ε) y = 0 1 y = y = 0 (x + 1)(ln |x + 1| + ε) 1 45) 2xy′ + y = 1 − x 1 1 HD gia ’ i: y′ + y = 2x 2x(1 − x) 11 www.VNMATH.com C y = √x √ √ x 1 x + 1 C′(x) = ⇒ C = ln |√ | + ε 2x(1 − x) 2 x − 1 √ 1 1 x + 1 y = √ ln | √ | + ε x 2 x − 1 46) xy′ − y = x2 sin x y HD gia ’ i: y′ − = x sin x y = Cx x y = (C − cos x)x 47) y′ cos2 x + y = tgx y(0) = 0 HD gia ’ i: → y = Ce−tgx; y = tgx − 1 ⇒ y = Ce−tgx + tgx − 1 y(0) = 0 ⇒ C = 1 y = tgx − 1 + e−tgx. √ 48) y′ 1 − x2 + y = arcsin x y(0) = 0 HD gia’i: y = Ce−arcsinx y = arcsinx − 1 ⇒
y = Ce−arcsinx + arcsinx − 1 y(0) = 0 ⇒ C = 1 ⇒
y = e−arcsinx + arcsinx − 1 1 49) y′ = 2x − y2 y(1) = 0 1 HD gia’i: x y′ = x′ 1 1 = ⇐⇒ x′ − 2x = −y2 x′ 2x − y2 x = Ce−2y y2 y 1 x = Ce−2y + − + 2 2 4 3 y(1) = 0 C = 43 y2 y 1 x = e−2y + − + 4 2 2 4 12 www.VNMATH.com z 50) y = x2 1 y∗ = ex 2
x2y′′ + 4xy′ + (x2 + 2)y = ex. z′x − 2z z′′x2 − 4z′x + 6z HD gia ’ i: y = zx2 =⇒ y′ = ; y′′ = x3 x4 ex z′′ + z = ex y∗ = 2 z = C1 cos x + C2 sin x cos x sin x ex y = C1 + C + x2 2 x2 2x2 51) yey = y′(y3 + 2xey) y(0) = −1 1 2 HD gia ’ i: x y′ = x′ − x = y2e−y x′ y C x = y C 1 C(y) = −e−y + C x = − y yey 1 C = e 52) y′ − y = cos x − sin x y x → ∞ HD gia’i: y = Cex + sin x y x → ∞ C = 0 53) y′ + sin y + x cos y + x = 0 π y(0) = 2 HD gia’i: y y y
y′ + sin y + x cos y + x = 0 ⇐⇒ y′ + 2 sin cos + x.2 cos2 = 0 2 2 2 y′ y ⇐⇒ y + tan + x = 0 2 cos2 2 2 y y′ z = tan =⇒ z′ = 2 y 2 cos2 2 z′ + z = −x z = 1 − x + Ce−x π y(0) = C = 0 y = 2 arctan(1 − x) 2 13 www.VNMATH.com x 54) y′ − x tan y = cos y HD gia ’ i: z = sin y, z′ − xz = x. 2 z = Cex2 − 1 2 sin y = z = Cex2 − 1 55) y′ − xy = x HD gia’i: 1 y = Ce x2 2 − 1 y √ 56) y′ + = x y. x HD gia’i: √ C 1 y = √ + x2. x 5 y 57) y′ − = x3 x HD gia’i: 1 y = Cx + x4. 3 58) y′ − y = y2. HD gia’i: 1 y2 = . Ce−2x − 1 y 59) y′ + = sin x x HD gia’i: C sin x y = + − cos x. x x 14 www.VNMATH.com √ 60) y′ − y = x y. HD gia’i: √ 1 y = Ce x 2 − x − 2. 61) y′ + 2xy = xe−x2 HD gia’i: x2 y = (C + )e−x2 2 y √ 62) y′ − 4 = x y. x HD gia’i: √ 1 y = ln x + Cx2. 2 63) y′ = y + 3x. 1 y” − y′ = x x y(x = 1) = 1 va` y′(x = 1) = 2. HD gia’i: R2. y′ y” − = x x x2 y = C1 + C2x + . 2 1 x2 y = − + x + . 2 2 64) y′ + ytgx = cos x HD gia’i: y = (C + x) cos x. 15 www.VNMATH.com y ex 65) y′ + = x( )y2. x ex + 1 HD gia’i: 1 y = . Cx − x ln(ex + 1) 66) (x + 1)y” + x(y′)2 = y′ HD gia’i: y′ = p x 6= −1 1 x p′ − p = − p2 (∗) x + 1 x + 1 z = p−1 6= 0 (∗) 1 x z′ + z = 1 + x x + 1 C z = x + 1 x2 + C 1 2(x + 1) z = 1 ⇒ y′ = = 2(x + 1) z x2 + C1 2 x ln + C |x2 + C1| + √ arctg √ 2 nˆe´u C1 > 0 C1 C1√ 1 x − −C1 ln ln |x2 + C1| + √ | √ | + C2 nˆe´u C1 < 0 −C1 x + −C1 y = C 67) x2y′ = y(x + y) 1 1
HD gia’i: x2y′ = y(x + y) ⇔ y′ − = y2 : y x2 1 1
z = y−1 (y 6= 0) : −z′ − z = . x x2 z = Cx 1 1 C(x) = ε − . z = x(ε − ) 2x2 2x2 2x y = εx2 − 1 68) yy” − (y′)2 = y3 1 y(0) = − 2 y′(0) = 0 16 www.VNMATH.com HD gia’i: y′ = p(y); y′′ = p.p′y dp py − p2 = y3, dy p(y) = y.z(y) dz 1 dy = ⇒ p z2 = 2(y + C = y |2y + C| dy z 1) ⇔ dx 1
y(0) = − ; y′(0) = 0 ⇒ C = 1 2 dy p |2y + 1| − 1 = yp|2y + 1| ⇒ ln = x + C2. dx p|2y + 1| + 1 1 y(0) = − ⇒ C2 = 0. 2 p |2y + 1| − 1 ln = x. p|2y + 1| + 1 √ 2y x 69) ydx + 2xdy = dy y(0) = π cos2 y 2 2 1 HD gia’i: x′ + x = .x 2 (∗) y cos2 y 1 1 1 z = x 2 z′ = x′ + x− 2 x′ (∗) 2 1 1 z′ + z = y cos2 y c z = y y C′ =
⇒ C(y) = ytgy + ln | cos y| + ε cos2 y 1 ε Z = tgy + ln | cos y| + y y 1 ε √ tgy + ln | cos y| + = x y y 1 √ y(0) = π ⇒ ε = 0 tgy + ln | cos y| = x y 70) xydy = (y2 + x)dx HD gia’i: y = 0 xy 1 y′ − y = y−1 z = y2 x 2
z′ − z = 2 → z = −2x + Cx2 x y2 = −2x + Cx2 √ 71) (y + xy)dx = xdy 17 www.VNMATH.com 1 1 HD gia’ 1 i: y′ − y = √ .y2 ; x 6= 0 x x 1 1 1 √ z = y2 : z′ − z = √ z = x(ln x + C) 2x x y = x(ln x + C)2 72) xy′ − 2x2√y = 4y √ 1 HD gia’i: z = y1−α = y ⇒ z′ = √ 2 y 4 z′ − z = 2x → z = Cx4 − x2 x y = (Cx2 − 1)2x4. 73) 2x2y′ = y2(2xy′ − y) HD gia’i: x y x′y3 − 2xy2 = −2x2 1 2z 2 z = z′ + = → y2 = x ln Cy2 x y y3 y = 0. 74) x2y′ = y(x + y) y(−2) = −4 HD gia’i: y(−2) = −4 y 6≡ 0 y2 1 1 y′ − 1y = z = y−1 z′ + z = − x2 x x2 z = Cx 1 2x C(x) = Cx − y = 2x Cx2 − 1 1 4x C = y = 2 x2 − 1 75) y′ − xy = −xy3 HD gia ’ i: y′ − xy = −xy3 y2(1 + Ce−x) = 1 76) xy′ + y = y2 ln x. HD gia ’ i: xy′ + y = y2 ln x 1 y = 1 + Cx + ln x y √ 77) y′ − 4 = x y x 18 www.VNMATH.com √ HD gia ’ i: z = y 2 x z′ − z = x 2 1 z = x2( ln |x| + C). 2 1 y = x4( ln |x| + C)2. 2 y 78) y′ + = y2xtgx. x HD gia’i: 1 y = Cx + xln|cosx| 79) 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 xy2 + 3y = C 80)
ex(2 + 2x − y2)dx − yexdy = 0 ∂P ∂Q HD gia ’ i: = = −2yex dex(2x −y2) = ∂y ∂x 0. ex(2x − y2) = C. 3 81)
(y2 + 1)2 dx + (y2 + 3xyp1 + y2)dy = 0 3 ∂P ∂Q
HD gia’i: p = (y2 + 1)2 ; Q = y2 + 3xyp1 + y2 ⇒ = = 3yp1 + y2 (∗) ∂y ∂x (∗) x y Z Z P (x, 0)dx + Q(x, y)dy = C 0 0 y3 3 ⇔ + x(1 + y2)2 = C 3 82)
(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 x y y2 R P (x, y R 0)dx + Q(x, y)dy = C ⇔ y sin x − cos2 x = C 2 x0=0 y0=0 83) (2x + 3x2y)dx = (3y2 − x3)dy HD gia’i: x2 + x3y − y3 = C x (x2 + 1) cos y 84) ( + 2)dx − dy = 0 sin y 2 sin2 y ∂P ∂Q x cos y HD gia’i: = = − ∂y ∂x sin2 y 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)
(y + ex sin y)dx + (x + ex cos y)dy = 0 HD gia’i: xy + ex sin y = C. 86)
(x + sin y)dx + (x cos y + sin y)dy = 0 HD gia’i: x2 + 2(x sin y − cos y) = C. x3 87) 3x2(1 + ln y)dx = (2y − )dy y HD gia’i: x3(1 + ln y) − y2 = C x3 88) 3x2(1 + ln y)dx = (2y − )dy y HD gia’i: x3(1 + ln y) − y2 = C 89)
(x + sin y)dx + (x cos y + sin y)dy = 0 HD gia’i: x2 + 2(x sin y − cos y) = C 20 www.VNMATH.com 90) 1 y2 x2 1 − dx + − dy = 0 x (x − y)2 (x − y)2 y x xy HD gia’i: ln + = C y x − y 91)
(sin xy + xy cos xy)dx + x2 cos xydy = 0 HD gia’i: x sin(xy) = C 92) (x + y2)dx − 2xydy = 0 1 HD gia ’ i: µ(x) = x2 2 y x = Ce x 93)
2xy ln ydx + (x2 + y2py2 + 1)dy = 0 1 HD gia’i: µ(y) = y 1 x2 ln y + (y2 3 +1) 2 = 0 3 94)
(x3 + xy2)dx + (x2y + y3)dy = 0 y(0) = 1 HD gia’i: x4 + 2x2y2 + y4 = C y(0) = 1 C = 1 95) a) − 2xydy + (y2 + x2)dx = 0 1 HD gia ’ i: µ(x) = x2 x2 − y2 = Cx.