《常微分方程》_(方道元_著)_课后习题答案__浙江大学出版社

(12)
dy (13) 1 + (1 + xy )exy + (1 + x2 exy ) dx = 0; dy dy (14) y sec2 x + sec x tan x + (2y + tan x) dx dx = 0;
dy ° 2yex + (y ° ex ) dx = 0;
(15) ydx ° (x2 + y 2 + x)dy = 0; (16) y (1 + xy )dx ° xdy = 0. (1) y¥0
!p ¯
+e
!p ¯
Z
!
q (t)e°
0
Rt
0
p(s) ds
dt
y (x) = y (x + ! ) , p ¯= 0
R
y (x) = y (x + ! ) , p ¯ 6= 0
p(x) dx
y=e
(C +
Z
q (x)e°
R
p(x) dx
dx),
C=
1 ¯ 1 ° e!p
Z
!
q (t)e°
0
Rt
0
p(s) ds
V1 +V2 V
t
[5m0 +
V1 +V2 m0 V1 t V 5 V1 +V2 (e
° 1)]. v0 v1
(x, y ) 8 > > > < 4. 200kg
4 3º
dx dt dy dt
= v0 = v1
150kg ∏∫c ∏∫s
> > > :
x(0) = 0, y (0) = 0 º ∫c ∫s ∏
dt
1. p dy (1) x dx ° 4xy = x2 y ;
dy dx dy dx
(2) (3)
°
= y2 +
xy 2(x2 °1)
1 4x2 ;
°
x 2y
= 0, y (0) = 1;
dy (4) x2 dx ° x2 y 2 = xy + 1;
(5)
dy dx dy dx
(6)
= 1 ° x + y 2 ° xy 2 ; = ex+y+3 ;
y tan x + sec x + y 2 = C C (15) y=0 y 6= 0 arctan C (16) y=0 y 6= 0
1 y2 1 y dx
xy x °y =C y
y x d(ln x y ) = ( x + y )dy
z = ln x y
°
x y 2 dy
+ xdx = 0
1 x + x2 y = Cy 2 C 2. dy x+y+1 = dx x°y+3 8
X +Y X °Y
arctan u ° ln C 3. u=
y °1 x+2
p
1 + u2 = ln |x| + C
dy ax + by + m = dx cx + dy + n a b c d m n ad ° bc 6= 0 dY aX + bY = dX cX + dY ad = bc ad ° bc 6= 0
1.
1000 / -20000 /
2
h
a = °2oooom/h2 8 > > > < > > > :
dh dt
v0 = 1000m/h, = v0 + at.
T
h(T ) = 0
dh dt |t=T
=0
1 v0 T + 2 aT 2 + C = 0,
2 h = v0 t + 1 2 at + C ,
(5) 2xydy ° (2y 2 ° x)dx = 0;
dy (7) 3xy 2 dx + y 3 + x3 = 0;
=
y x+y 3 .
3
(1)
e°x y = Cex ° C
R
1 1°x2
sin x + cos x 2
(2)
e
°
dx
C (3) x
y 0 = ex + y (x)
1+x º (C + ) 1°x 2 q 1+x C =1° º y= 1 2 °x y= y (0) = 1
8 < Vc (t) = : V (t) =
s
Mc mc ∏ (1 Ms ms ∏ (1
° e° mc t )
∏
∏
° e° ms t )
1. (1)
dy dx
d y dy (3) x2 dx 2 ° 2x dx + y = 2x sin x; d y x dy (5) y dx 3 ° e dx + 3xy = 0;
dx
z = x2 (C + x°1 y 2 = x2 (C + x°1 ) (6) y¥0 C y 6= 0 °y °2 dz 2z ln x = ° dx x x e°
R
2 x
z = y °1
dx
y °1 = x2 (C + 4
1 + 2 ln x ) 4x2
C (7) y=0 e
R
1 x
y 6= 0
(3) y = C1 e2x + C2 e°2x , (4) y = cxex , (5) y = ecx ,
d y dx2
2
dy 1 (2) y = ° x , x2 dx ° x2 y 2 ° xy ° 1 = 0; dy dx2
= y 2 ° (x2 + 1)y + 2x; ° 4y = 0( c c C1 C2 ); ); );
dy (7) cos y sin x dx = sin y cos x; dy (8) 2xy dx = 3y 2 ° x2 ; p dy (9) (x ° xy ) dx = y;
(10) e y
y2 2
x
dy dx
+ y (1 + e y ) = 0;
x
dy (11) 2x sin y + y 3 ex + (x2 cos y + 3y 2 ex ) dx = 0;
p(x) q (x) (1) q (x) ¥ 0
! (2.4.23) ! Z 1 ! p ¯= p(x) dx = 0 ! 0 5 p(x)
(2) q (x) 6= 0
(2.4.23)
!
p(x)
p ¯ 6= 0
y (x) = y (x + ! ) , (1) q (t) = 0 (2) q (t) 6= 0 C = Ce
dx
y 0 = ° 31 xy ° C x3 ° x 4
x2 3y 2
3y 2
z = y3
y= C
d(xy 3 ) dx
(8)
y¥0
= °x3
y 6= 0
ydx°xdy y2
° ydy = 0
x y2 ° =C y 2 C 2. y = '(x) dy + a(x)y ∑ 0, (x ∏ 0). dx '(x) ∑ '(0)e° e 3. f (x) (°1, +1) dy ° y = f (x) dx (°1, +1) ! |f (x)| ∑ M R ex f (s)e°s ds e°x (°1, +1) y = ex (C + R f (s)e°s ds) C f (x) !
p
1 ° x2 )
1 1 ° Cx ° x ln |x| 2x y= u0 +
u x
C u=y+
1 2x
xy = ° 1 2 u
° u2 = 0
1 Cx°x ln |x|
°
1 2x
(4) xy = u ln |x| +
u0 = =C
u2 +2u+1 x
u2 + 2u + 1 = 0 xy = °1
1 xy +1
x
x2 sin y + y 3 ex = C C (12) [ y2 x e ° 2ye2x ]dx + (ex y ° e2x )dy = 0 2 y2 x e ° ye2x = C 2 C (13) dy + dx + {x2 exy dy + [(1 + xy )exy ]dx} = 0
y + x + xexy = C C (14) (y sec2 x)dx + tan xdy + (sec x tan x)dx + 2ydy = 0
1
mc , m s Vc , V s , 8 < mc g ° ∏Vc = mc dVc + ΩVc g dt : m g ° ∏V = m dVs + ΩV g
s s s dt s
Mc Ms
=
4 3
8 < mc g ° ΩVc g = Mc : m g ° ΩV g = M
s s
s
mc ms
=
4 3
dy 2 ( dx )
dy ° 2 dx + y = 0(
°
d2 y y dx 2
= 0(
2
(6) y =
8 C1 )2 > ° (x°4 , °1 < x < C1 ; > > < > > > : 0, C1 < x < C2 ,
(x°C1 )2 , 4
dy dx
=
C2 < x < +1,
p |y |.
1 3 Y +k (a) h k dy x+y+1 = dx x°y+3 dY X +Y = dX X °Y dy x+y+1 = dx x°y+3 (1) x = X + h, y = Y + k
dY dX
x = X + h, y =
(b)
=
(2)
h = °2, k = 1
8 < h+k+1=0 : h°k+3=0
(1) (2) (3) (4) (5) (6) 3. (1) (2) (3) (4)
y = Cx + x2 xy = C
y=9 2C +
C 2 xx
x (1)y 00 = 2 (2)y + xy 0 = 0 (3)y 000 + (y 0 )2 y 000 = 3y 0 (y 00 )2 (4)y 000 = 0
3
= 4x3 ° y sin x;
2
(2) (4) (6)
dy dx2 dy dx
3
+ cos y + 4x = 0;
dy + 3 dx ° 6y = 0.
dy 2 ° ( dx ) + 2xy ;
d y dx3
(1) (2) (3) (4) (5) (6) 2. (1) y = 1 + x2 ,
dy dx
r
e°x
y = ex (C + x) C (4) 3y 2 z = y3 dz 3z = 3x3 + dx x e°
R
3 x
C=1
y = ex (1 + x)
dx
z = x3 (C + 3x) C (5) z = y 2 dz 2z = °1 dx x e°
R
2 x
y 3 = x3 (C + 3x)
y 6= 0
1 °1 2 2y
z = y2
1
e°2x
y = (Ce2x ° 6
x 1 2 ° ) 4 8
C (2) 2y z = y2 y= C (3) z = xy z0 =
4z 2 +4z +1 4x
e p C
°
R
x x2 °1
dx
1 ° x2 (C °
z = °1 2 y=
p y = 2 1 ° x2 ° (1 ° x2 ) z 6= ° 1 2
Rx
0
Rx
0
a(t) dt
, (x ∏ 0).
a(s) ds
|y | ∑ Cex + M
f (x + ! ) = f (x)
! )e°s ds = y (x) 4.
C =0 y = R R y (x + ! ) = e(x+!) f (s + ! )e°(s+!) ds = e(x+!) e°! f (s + dy p(x)y + q (x), dx
3 z0 = x z°x
e°
R
3 x
dx
y 2 = |x|3 (C + |x|°1 ) C (9) u =
y x
u
x2 (1 °
p
p u)u0 = xu u
u=0 px
y
y=0
u 6= 0
y = Ce°2 C (10) u =
x y u +e y du dy = ° 1+eu
u
x + ye y = C C (11) [(x2 cos y )dy + (2x sin y )dx] + [(y 3 ex )dx + (3y 2 ex )dy ] = 0 7
0 T = °v a
C
h(0) = C = 25m. V V2 / A V 1 + V2 2000 / V1 2000 / A A
2.
A 5m0
m0 5
A A P(t), 8 < V P 0 (t) + P (t)(V1 + V2 ) = : P (0) = 5m
0 m0 5 V1
P (t) = e° 3.
(5) (6) (7)
e°y + ex+3 = C sin y = 0 C y = k º (k 2 Z)
2 y = tan(x ° 1 2 x + C)
ln |x| +
xy = °1
1 xy +1
=C C
C
xy 6= °1
C sin y 6= 0
(8) z = y 2
sin2 y ° C sin2 x = 0
an°mc bc°ad , X
ax + by + m = 0 =x° µ
nb°md ad°bc dY dX
cx + dy + n = 0
aX +bY cX +dY dy dx
Y =y° ad = bc C
wenku.baidu.com
=
°md an°mc ( nb ad°bc , bc°ad )
1. (1) (2)
dy dx d y dx2
2
= y + sin x; °
1 1°x2 y = 1 + x, Rx ex + 0 y (t) dt; x4 +y 3 xy 2 ;
y (0) = 1;
(3) y = (4)
dy dx
=
(6) (y ln x ° 2)ydx = xdy ; (8)
dy dx