Differential Equations Solutions
Seatwork
Homework 6-9
October 19, 2025
Seatwork 6: Homogeneous Coefficients
1. (x − 2y)dx + (2x + y)dy = 0
Let y = vx, dy = vdx + xdv.
(x − 2vx)dx + (2x + vx)(vdx + xdv) = 0
x(1 − 2v)dx + x(2 + v)(vdx + xdv) = 0
(1 − 2v)dx + (2v + v 2 )dx + x(2 + v)dv = 0
(1 + v 2 )dx + x(2 + v)dv = 0
v+2
dx
+ 2
dv = 0
x Z v +1
Z
Z
dx
v
2
+
dv
+
dv = ln |C|
x
v2 + 1
v2 + 1
1
ln |x| + ln(v 2 + 1) + 2 arctan(v) = ln |C|
2
y
1
y2
ln |x| + ln
+
1
+
2
arctan
= ln |C|
2
x2
x
2
y
1
x + y2
ln |x| + ln
= ln |C|
+ 2 arctan
2
x2
x
y
1
= ln |C|
ln |x| + ln(x2 + y 2 ) − ln |x| + 2 arctan
2
x
y
1
ln(x2 + y 2 ) + 2 arctan
= ln |C|
2
x
y
ln(x2 + y 2 ) + 4 arctan
= C1
x
1
2. (x2 + y 2 )dx + xy dy = 0
Let y = vx, dy = vdx + xdv.
(x2 + v 2 x2 )dx + x(vx)(vdx + xdv) = 0
x2 (1 + v 2 )dx + vx2 (vdx + xdv) = 0
(1 + v 2 )dx + (v 2 )dx + xv dv = 0
(1 + 2v 2 )dx + xv dv = 0
dx
v
+
dv = 0
x Z 1 + 2v 2
Z
dx 1
4v
+
dv = C
x
4
1 + 2v 2
1
ln |x| + ln(1 + 2v 2 ) = C
4
4 ln |x| + ln(1 + 2(y/x)2 ) = C1
2
x + 2y 2
4
ln(x ) + ln
= C1
x2
2
2
4 x + 2y
ln x ·
= C1
x2
x2 (x2 + 2y 2 ) = C
3. (5v − u)du + (3v − 7u)dv = 0
Let u = zv, du = zdv + vdz.
(5v − zv)(zdv + vdz) + (3v − 7zv)dv = 0
v(5 − z)(zdv + vdz) + v(3 − 7z)dv = 0
(5z − z 2 )dv + v(5 − z)dz + (3 − 7z)dv = 0
(−z 2 − 2z + 3)dv + v(5 − z)dz = 0
5−z
dv
z−5
dv
+ 2
dz = 0 =⇒
−
dz = 0
v
z + 2z − 3
v
(z + 3)(z − 1)
z−5
A
B
Partial Fractions:
=
+
=⇒ A = 2, B = −1
(z + 3)(z − 1)
z+3 z−1
Z
Z dv
2
1
−
−
dz = C
v
z+3 z−1
ln |v| − 2 ln |z + 3| + ln |z − 1| = C
v(z − 1)
v(z − 1)
ln
= C =⇒
= C1
2
(z + 3)
(z + 3)2
v(u/v − 1)
v(u − v)/v
=
C
=⇒
= C1
1
(u/v + 3)2
((u + 3v)/v)2
u−v
v 2 (u − v)
= C1 =⇒
= C1
(u + 3v)2 /v 2
(u + 3v)2
(u − v) = C(u + 3v)2
*Note: The C constant can be on either side. (u + 3v)2 = C(u − v) is also correct.*
2
Homework 6: Homogeneous Coefficients
x+3y
dy
= 3x+y
1. dx
dy
dv
= v + x dx
.
Let y = vx, dx
dv
x + 3vx
1 + 3v
=
=
dx
3x + vx
3+v
dv
1 + 3v
1 + 3v − 3v − v 2
1 − v2
x
=
−v =
=
dx
3+v
3+v
3+v
3+v
3+v
dx
dx
=⇒
dv =
dv =
1 − v2
x
(1 − v)(1 + v)
x
A
B
+
=⇒ A = 2, B = 1
Partial Fractions:
1 −Zv 1 + v
Z 2
dx
1
+
dv =
1−v 1+v
x
−2 ln |1 − v| + ln |1 + v| = ln |x| + C
1+v
1+v
= ln |x| + C =⇒
= C1 x
ln
2
(1 − v)
(1 − v)2
(x + y)/x
1 + y/x
= C1 x =⇒
= C1 x
2
(1 − y/x)
((x − y)/x)2
x(x + y)
= C1 x =⇒ x + y = C(x − y)2
(x − y)2
v+x
2. [x csc( xy ) − y]dx + x dy = 0
Let y = vx, dy = vdx + xdv.
(x csc(v) − vx)dx + x(vdx + xdv) = 0
x(csc(v) − v)dx + xv dx + x2 dv = 0
(csc(v) − v + v)dx + x dv = 0
csc(v)dx + x dv = 0
Z
Z
dx
dv
dx
+
= 0 =⇒
+ sin(v)dv = C
x
csc(v)
x
ln |x| − cos(v) = C
ln |x| − cos(y/x) = C
3
x
3. y 2 dy = x(x dy − y dx)e y
Rearrange: (xyex/y )dx + (y 2 − x2 ex/y )dy = 0. Let x = vy, dx = vdy + ydv.
(vy · yev )(vdy + ydv) + (y 2 − (vy)2 ev )dy = 0
(vy 2 ev )(vdy + ydv) + y 2 (1 − v 2 ev )dy = 0
(v 2 ev )dy + (vyev )dv + (1 − v 2 ev )dy = 0
(v 2 ev + 1 − v 2 ev )dy + vyev dv = 0
dy + vyev dv = 0
dy
+ vev dv = 0
y
Z
Z
Z
dy
v
+ ve dv = C (Integration by parts:
vev dv = vev − ev )
y
ln |y| + vev − ev = C
x x/y
x
x/y
x/y
ln |y| + e − e = C =⇒ ln |y| + e
−1 =C
y
y
3
dy
4. dx
− xy = xy 3 , y(1) = 3
dy
This is homogeneous. dx
= xy +
y 3
dy
dv
. Let y = vx, dx
= v + x dx
.
x
dv
= v + v3
dx
dv
dx
dv
= v 3 =⇒ 3 =
x
dx Z
v
x
Z
dx
v −3 dv =
x
−2
v
1
= ln |x| + C =⇒ − 2 = ln |x| + C
−2
2v
1
x2
−
= ln |x| + C =⇒ − 2 = ln |x| + C
2(y/x)2
2y
2
1
1
= ln |1| + C =⇒ − = C
Apply y(1) = 3 : −
2
2(3 )
18
2
x
1
− 2 = ln |x| −
2y
18
2
1
x
1 − 18 ln |x|
=
− ln |x| =
2
2y
18
18
2
2
9x = y (1 − 18 ln |x|)
v+x
4
y
y
5. (x + ye x )dx − xe x dy = 0, y(1) = 0
Let y = vx, dy = vdx + xdv.
(x + vxev )dx − xev (vdx + xdv) = 0
x(1 + vev )dx − vxev dx − x2 ev dv = 0
(1 + vev − vev )dx − xev dv = 0
dx − xev dv = 0
dx
− ev dv = 0
x
Z
Z
dx
− ev dv = C
x
ln |x| − ev = C =⇒ ln |x| − ey/x = C
Apply y(1) = 0 : ln |1| − e0/1 = C =⇒ 0 − 1 = C =⇒ C = −1
ln |x| − ey/x = −1 =⇒ ln |x| + 1 = ey/x
Seatwork 7: Exact Equations
1. (2x − 3y)dx + (2y − 3x)dy = 0
M = 2x − 3y, N = 2y − 3x. ∂M
= −3, ∂N
= −3. It is exact.
∂y
∂x
Z
F =
(2x − 3y)dx = x2 − 3xy + g(y)
∂F
= −3x + g ′ (y) = N = 2y − 3x
∂y
′
g (y) = 2y =⇒ g(y) = y 2
Solution: x2 − 3xy + y 2 = C
2. (cos 2y − 3x2 y 2 )dx + (cos 2y − 2x sin 2y − 2x3 y)dy = 0
M = cos 2y − 3x2 y 2 , N = cos 2y − 2x sin 2y − 2x3 y. ∂M
= −2 sin 2y − 6x2 y.
∂y
−2 sin 2y − 6x2 y. It is exact.
Z
F = (cos 2y − 3x2 y 2 )dx = x cos 2y − x3 y 2 + g(y)
∂F
= −2x sin 2y − 2x3 y + g ′ (y) = N
∂y
1
g ′ (y) = cos 2y =⇒ g(y) = sin 2y
2
1
Solution: x cos 2y − x3 y 2 + sin 2y = C
2
5
∂N
∂x
=
3. (r + sin θ − cos θ)dr + r(sin θ + cos θ)dθ = 0
= cos θ + sin θ. ∂N
= sin θ + cos θ. It is
M = r + sin θ − cos θ, N = r sin θ + r cos θ. ∂M
∂θ
∂r
exact.
Z
F = (r sin θ + r cos θ)dθ = −r cos θ + r sin θ + h(r)
∂F
= − cos θ + sin θ + h′ (r) = M = r + sin θ − cos θ
∂r
1
′
h (r) = r =⇒ h(r) = r2
2
1 2
Solution: r + r sin θ − r cos θ = C
2
Homework 7: Exact Equations
1. (sin x sin y + tan x)dx − cos x cos y dy = 0
M = sin x sin y + tan x, N = − cos x cos y. ∂M
= sin x cos y. ∂N
= −(− sin x) cos y =
∂y
∂x
sin x cos y. It is exact.
Z
F = (− cos x cos y)dy = − cos x sin y + h(x)
∂F
= −(− sin x) sin y + h′ (x) = sin x sin y + h′ (x) = M
∂x
Z
h′ (x) = tan x =⇒ h(x) =
tan xdx = ln | sec x|
Solution: − cos x sin y + ln | sec x| = C
2. (2xy − tan y)dx + (x2 − x sec2 y)dy = 0
M = 2xy − tan y, N = x2 − x sec2 y. ∂M
= 2x − sec2 y. ∂N
= 2x − sec2 y. It is exact.
∂y
∂x
Z
F =
(2xy − tan y)dx = x2 y − x tan y + g(y)
∂F
= x2 − x sec2 y + g ′ (y) = N
∂y
′
g (y) = 0 =⇒ g(y) = 0
Solution: x2 y − x tan y = C
6
y
t
y
3. ( 1t + t12 − t2 +y
2 )dt + (ye + t2 +y 2 )dy = 0
y
t
y
+ t2 +y
M = 1t + t12 − t2 +y
2 , N = ye
2.
∂M
∂y
2
2
2
2
−t
= − (t +y(t2)(1)−y(2y)
= (ty2 +y
2 )2 .
+y 2 )2
∂N
∂t
=
2 −t2
(t2 +y 2 )(1)−t(2t)
= (ty2 +y
2 )2 . It is exact.
(t2 +y 2 )2
Z t
y
ye + 2
dy
F =
t + y2
Z
Z
yey dy = yey − ey
t
(by parts)
1
arctan(y/t) = arctan(y/t)
t
F = yey − ey + arctan(y/t) + h(t)
y
1
−y/t2
∂F
′
=
+
h
(t)
=
·
−
+ h′ (t)
2
2
2
2
2
∂t
1 + (y/t)
t
(t + y )/t
−y
= 2
+ h′ (t) = M
t + y2
1
1
1
h′ (t) = + 2 =⇒ h(t) = ln |t| −
t t
t
1
Solution: ln |t| − + yey − ey + arctan(y/t) = C
t
t2 + y 2
dy = t ·
4. (x + y)2 dx + (2xy + x2 − 1)dy = 0, y(1) = 1
= 2x + 2y. ∂N
= 2y + 2x. It is
M = (x + y)2 = x2 + 2xy + y 2 , N = 2xy + x2 − 1. ∂M
∂y
∂x
exact.
Z
F = (2xy + x2 − 1)dy = xy 2 + x2 y − y + h(x)
∂F
= y 2 + 2xy + h′ (x) = M = x2 + 2xy + y 2
∂x
1
h′ (x) = x2 =⇒ h(x) = x3
3
1
General Solution: xy 2 + x2 y − y + x3 = C
3
1
Apply y(1) = 1 :(1)(1)2 + (1)2 (1) − 1 + (1)3 = C
3
1
4
1 + 1 − 1 + = C =⇒ C =
3
3
1
4
Solution: xy 2 + x2 y − y + x3 =
=⇒ 3xy 2 + 3x2 y − 3y + x3 = 4
3
3
7
5. (y 2 cos x − 3x2 y − 2x)dx + (2y sin x − x3 + ln y)dy = 0, y(0) = e
= 2y cos x−3x2 . ∂N
= 2y cos x−3x2 .
M = y 2 cos x−3x2 y−2x, N = 2y sin x−x3 +ln y. ∂M
∂y
∂x
It is exact.
Z
F = (y 2 cos x − 3x2 y − 2x)dx = y 2 sin x − x3 y − x2 + g(y)
∂F
= 2y sin x − x3 + g ′ (y) = N
∂y
Z
′
g (y) = ln y =⇒ g(y) = ln ydy = y ln y − y
(by parts)
General Solution: y 2 sin x − x3 y − x2 + y ln y − y = C
Apply y(0) = e :(e2 sin 0) − (0)3 (e) − 02 + (e ln e) − e = C
0 − 0 − 0 + e(1) − e = C =⇒ C = 0
Solution: y 2 sin x − x3 y − x2 + y ln y − y = 0
Seatwork 8: Linear Equations
1. y ′ = x − 2y cot 2x
y ′ + (2 cot 2x)y = x. This is Linear. P (x) = 2 cot 2x, Q(x) = x.
Z
Z
cos 2x
P (x)dx = 2
dx = ln | sin 2x|
sin 2x
IF = I = eln | sin 2x| = sin 2x
Z
Z
y · I = I · Qdx =⇒ y sin 2x = x sin 2xdx
Z
Z
1
1
x sin 2xdx = x(− cos 2x) − (− cos 2x)dx (by parts)
2
2
Z
x
1
x
1
= − cos 2x +
cos 2xdx = − cos 2x + sin 2x + C
2
2
2
4
x
1
y sin 2x = − cos 2x + sin 2x + C
2
4
x
1
y = − cot 2x + + C csc 2x
2
4
8
2. 2y dx = (x2 − 1)(dx − dy)
2y dx = (x2 − 1)dx − (x2 − 1)dy
(x2 − 1)dy = (x2 − 1 − 2y)dx
dy
x2 − 1 − 2y
2y
=
=1− 2
2
x −1
x −1
dx
2
2
dy
y = 1. P (x) = 2
, Q(x) = 1.
+
dx
x2 − 1
x −1
Z
Z
Z 2
1
1
P (x)dx =
dx =
−
dx
(x − 1)(x + 1)
x−1 x+1
x−1
= ln |x − 1| − ln |x + 1| = ln
x+1
x−1
x−1
IF = I = eln | x+1 | =
x+1 Z
Z
x−1
x−1
y · I = I · Qdx =⇒ y
=
dx
x+1
x+1
Z Z
2
(x + 1) − 2
dx =
1−
dx = x − 2 ln |x + 1| + C
x+1
x+1
x−1
y
= x − 2 ln |x + 1| + C
x+1
3. y ′ = 1 + 3y tan x
y ′ − (3 tan x)y = 1. P (x) = −3 tan x, Q(x) = 1.
Z
Z
sin x
P (x)dx = −3
dx = 3 ln | cos x| = ln | cos3 x|
cos x
3
IF = I = eln | cos x| = cos3 x
Z
Z
3
y · I = I · Qdx =⇒ y cos x = cos3 xdx
Z
Z
3
cos xdx = (1 − sin2 x) cos xdx (Let u = sin x)
Z
u3
sin3 x
= (1 − u2 )du = u −
+ C = sin x −
+C
3
3
1
y cos3 x = sin x − sin3 x + C
3
9
Homework 8: Linear Equations
1. xy ′ + (1 + x)y = e−x sin 2x
−x sin 2x
y=e
y ′ + 1+x
x
−x
2x
. P (x) = x1 + 1, Q(x) = e sin
.
x
Z
Z 1
+ 1 dx = ln |x| + x
P (x)dx =
x
x
IF = I = eln |x|+x = eln |x| ex = xex
−x
Z
Z
e sin 2x
x
x
y · I = I · Qdx =⇒ yxe = (xe )
dx
x
Z
1
x
yxe = sin 2xdx = − cos 2x + C
2
x
2yxe + cos 2x = C1
dy
+ y = ln x
2. (x + 1) dx
dy
1
ln x
1
ln x
+ x+1
y = x+1
. P (x) = x+1
, Q(x) = x+1
.
dx
Z
Z
1
dx = ln |x + 1|
x+1
IF = I = eln |x+1| = x + 1
Z
Z
ln x
y · I = I · Qdx =⇒ y(x + 1) = (x + 1)
dx
x+1
Z
Z
1
y(x + 1) = ln xdx = x ln x − x · dx (by parts)
x
Z
= x ln x − dx = x ln x − x + C
P (x)dx =
y(x + 1) = x ln x − x + C
dr
+ r sec θ = cos θ
3. dθ
This is Linear. P (θ) = sec θ, Q(θ) = cos θ.
Z
Z
P (θ)dθ = sec θdθ = ln | sec θ + tan θ|
IF = I = eln | sec θ+tan θ| = sec θ + tan θ
Z
Z
r · I = I · Qdθ =⇒ r(sec θ + tan θ) = (sec θ + tan θ) cos θdθ
Z Z
1
sin θ
r(sec θ + tan θ) =
+
cos θdθ = (1 + sin θ)dθ
cos θ cos θ
r(sec θ + tan θ) = θ − cos θ + C
10
di
4. L dt
+ Ri = E sin(wt), i(0) = 0
di
+R
i = EL sin(wt). P (t) = R/L, Q(t) = EL sin(wt).
dt
L
Z
R
R
t =⇒ IF = I = e L t
L
Z
Z
R
R
E
t
t
L
L
sin(wt) dt
i · I = I · Qdt =⇒ ie = e
L
Z
eat
Use integral
eat sin(bt)dt = 2
(a sin(bt) − b cos(bt))
a + b2
R2
R2 + w2 L2
Here a = R/L, b = w.a2 + b2 = 2 + w2 =
L2
" L R
#
t
L
R
e
R
E
sin(wt) − w cos(wt)
+C
ie L t =
L (R2 + w2 L2 )/L2 L
"
#
t
2 R
L
R
E
L
e
R
sin(wt)
−
wL
cos(wt)
ie L t =
+C
L R 2 + w 2 L2
L
P (t)dt =
R
Ee L t
ie = 2
(R sin(wt) − wL cos(wt)) + C
R + w2 L2
R
E
(R sin(wt) − wL cos(wt)) + Ce− L t
i(t) = 2
2
2
R +w L
E
EwL
Apply i(0) = 0 :0 = 2
(0 − wL) + C =⇒ C = 2
2
2
R +w L
R + w 2 L2
R
EwL
E
(R sin(wt) − wL cos(wt)) + 2
e− L t
Solution: i(t) = 2
2
2
2
2
R +w L
R +w L
R
t
L
5. xy ′ + y = ex , y(−1) = 4
d
(xy) = ex .
This is an exact linear equation: dx
Z
Z
d
(xy)dx = ex dx
dx
xy = ex + C
Apply y(−1) = 4 :(−1)(4) = e−1 + C =⇒ −4 =
C = −4 − e−1
Solution: xy = ex − 4 − e−1
11
1
+C
e
Seatwork 9: Integrating Factors by Inspection
1. y(x4 − y 2 )dx + x(x4 + y 2 )dy = 0
x4 ydx − y 3 dx + x5 dy + xy 2 dy = 0
(x4 ydx + x5 dy) + (xy 2 dy − y 3 dx) = 0
x4 (ydx + xdy) + y 2 (xdy − ydx) = 0
xdy − ydx
4
2
2
x d(xy) + y · x ·
=0
x2
x4 d(xy) + (xy)2 d(y/x) = 0 (Let u = xy, v = y/x)
x4 = (u/v)2 .(u/v)2 du + u2 dv = 0
(1/v 2 )du + dv = 0 (This is not exact, needs IF v 2 )
Multiply by v 2 = (y/x)2 :d(u) + v 2 d(v) = 0
Z
Z
d(xy) + (y/x)2 d(y/x) = C
1 y 3
xy +
=C
3 x
3x4 y + y 3 = C1 x3
2. y(x3 exy − y)dx + x(y + x3 exy )dy = 0
x3 yexy dx − y 2 dx + xydy + x4 exy dy = 0
(x3 yexy dx + x4 exy dy) + (xydy − y 2 dx) = 0
x3 exy (ydx + xdy) + y(xdy − ydx) = 0
ydx − xdy
3 xy
2
=0
x e d(xy) − y · y ·
y2
x3 exy d(xy) − y 3 d(x/y) = 0 (Let u = xy, v = x/y)
p
√
x = uv, y = u/v.x3 = (uv)3/2 , y 3 = (u/v)3/2 .
(uv)3/2 eu du − (u/v)3/2 dv = 0
u3/2 v 3/2 eu du = u3/2 v −3/2 dv
v 3 eu du = dv =⇒ eu du = v −3 dv
Z
Z
u
e du = v −3 dv
v −2
+C
−2
1
1
exy = − 2 + C =⇒ exy = −
+C
2v
2(x/y)2
y2
y2
exy = − 2 + C =⇒ exy + 2 = C
2x
2x
eu =
12
3. y(3x3 − x + y)dx + x2 (1 − x2 )dy = 0
(3x3 y − xy + y 2 )dx + (x2 − x4 )dy = 0
y 2 dx + x2 dy − x4 dy + 3x3 ydx − xydx = 0
(y 2 dx + x2 dy) − (xydx + x4 dy) + 3x3 ydx = 0
y2
y
Divide by x2 : 2 dx + dy − x2 dy + 3xydx − dx = 0
x
x
2
y
y
dx + 2 dx = 0
(1 − x2 )dy + 3xy −
x
x
2
1
y
(1 − x2 )y ′ + y 3x −
= − 2 (Bernoulli)
x
x
y2
1
−1 ′
−2 ′
2 2 ′
=− 2
Let v = y .v = −y y . − (1 − x )y v + y 3x −
x
x
3x − 1/x
1
(Linear in v)
v′ − v
=
1 − x2
x2 (1 − x2 )
Z
Z 3x2 − 1
1
1
1
P (x) = −
. P (x)dx =
−
+
dx
x(1 − x2 )
x 1−x 1+x
= ln |x| + ln |1 − x| + ln |1 + x| = ln |x(1 − x2 )|
IF = I = x(1 − x2 )
Z
Z
1
1
2
2
dx
=
dx
v · x(1 − x ) = x(1 − x ) · 2
x (1 − x2 )
x
vx(1 − x2 ) = ln |x| + C
1
(x − x3 ) = ln |x| + C
y
y(ln |x| + C) = x − x3
13
Homework 9: Integrating Factors by Inspection
1. (xn y n+1 + ay)dx + (xn+1 y n + bx)dy = 0
xn y n (ydx) + aydx + xn y n (xdy) + bxdy = 0
xn y n (ydx + xdy) + (aydx + bxdy) = 0
(xy)n d(xy) + aydx + bxdy = 0
Try IF = xp y q . Requires p = q and a(q + 1) = b(p + 1).
If a ̸= b, then p + 1 = 0 =⇒ p = −1. So p = q = −1.
a
b
n−1 n
n n−1
IF = 1/(xy). x y +
dx + x y
+
dy = 0
x
y
M = xn−1 y n + a/x,N = xn y n−1 + b/y
∂M
∂N
= nxn−1 y n−1 .
= nxn−1 y n−1 . It is exact.
∂y
∂x
Z
xn y n
n−1 n
F = (x y + a/x)dx =
+ a ln |x| + g(y)
n
xn · ny n−1
∂F
=
+ g ′ (y) = xn y n−1 + g ′ (y) = N
∂y
n
′
g (y) = b/y =⇒ g(y) = b ln |y|
(xy)n
Solution (if n ̸= 0) :
+ a ln |x| + b ln |y| = C
n
2. x4 y ′ = −x3 y − csc(xy)
dy
= −x3 y − csc(xy)
dx
x4 dy = (−x3 y − csc(xy))dx
(x3 y + csc(xy))dx + x4 dy = 0
(x3 ydx + x4 dy) + csc(xy)dx = 0
x3 (ydx + xdy) + csc(xy)dx = 0
x3 d(xy) + csc(xy)dx = 0 (Let u = xy)
x3 du + csc(u)dx = 0
du
du
csc(u)
+ csc(u) = 0 =⇒
=− 3
x3
dx
dx
Z x
Z
du
−3
= −x dx =⇒
sin(u)du = −x−3 dx
csc(u)
x−2
1
− cos(u) = −
+C = 2 +C
−2
2x
1
− cos(xy) = 2 + C
2x
1
cos(xy) + 2 = C1
2x
x4
14
3. y(x3 y 3 + 2x2 − y)dx + x3 (xy 3 − 2)dy = 0
(x3 y 4 + 2x2 y − y 2 )dx + (x4 y 3 − 2x3 )dy = 0
Try IF = 1/(xy)2 = x−2 y −2 .
M = xy 2 + 2/y − 1/x2 ,N = x2 y − 2x/y 2
∂M
∂N
= 2xy − 2/y 2 .
= 2xy − 2/y 2 . It is exact.
∂y
∂x
Z
x2 y 2 2x 1
F = (xy 2 + 2/y − 1/x2 )dx =
+
+ + g(y)
2
y
x
2
x · 2y 2x
∂F
2x
=
− 2 + g ′ (y) = x2 y − 2 + g ′ (y) = N
∂y
2
y
y
′
g (y) = 0 =⇒ g(y) = 0
x2 y 2 2x 1
Solution:
+
+ =C
2
y
x
4. xy(y 2 + 1)dx + (x2 y 2 − 2)dy = 0, y(1) = 1
(xy 3 + xy)dx + (x2 y 2 − 2)dy = 0
(xy 3 dx + x2 y 2 dy) + xydx − 2dy = 0
xy 2 (ydx + xdy) + xydx − 2dy = 0
xy 2 d(xy) + xydx − 2dy = 0 (Let u = xy =⇒ y = u/x)
x(u/x)2 du + udx − 2dy = 0
xdu − udx
2
(u /x)du + udx − 2
=0
x2
2
2
u
2u
−
du + u + 2 dx = 0
x
x
x
2
2
u −2
du + u 1 + 2 dx = 0
x
x
2
u −2
2
du + x 1 + 2 dx = 0
u
x
Z
Z
(u − 2/u)du + (x + 2/x)dx = C
u2
x2
− 2 ln |u| +
+ 2 ln |x| = C
2
2
(xy)2
x2
− 2 ln |xy| +
+ 2 ln |x| = C
2
2
x2 y 2
x2
− 2 ln |x| − 2 ln |y| +
+ 2 ln |x| = C
2
2
x2 y 2 + x 2
− 2 ln |y| = C =⇒ x2 y 2 + x2 − 4 ln |y| = C1
2
Apply y(1) = 1 :(1)2 (1)2 + (1)2 − 4 ln |1| = C1 =⇒ 1 + 1 − 0 = C
Solution: x2 y 2 + x2 − 4 ln |y| = 2
15
5. y(x2 + y)dx + x(x2 − 2y)dy = 0, y(1) = 2
(x2 y + y 2 )dx + (x3 − 2xy)dy = 0
Test for IF xa y b . Equating partials requires:
(b + 1)xa+2 y b + (b + 2)xa y b+1 = (a + 3)xa+2 y b − 2(a + 1)xa y b+1
1) b + 1 = a + 3 =⇒ a = b − 2
2) b + 2 = −2(a + 1) = −2(b − 2 + 1) = −2(b − 1) = −2b + 2
3b = 0 =⇒ b = 0. =⇒ a = −2. IF = x−2 .
(y + y 2 /x2 )dx + (x − 2y/x)dy = 0
M = y + y 2 /x2 ,N = x − 2y/x
∂M
∂N
= 1 + 2y/x2 .
= 1 − 2y(−1/x2 ) = 1 + 2y/x2 . Exact
∂y
∂x
Z
2 y2
+ h(x) =
F = (x − 2y/x)dy = xy −
x 2
∂F
= y − y 2 (−1/x2 ) + h′ (x) = y + y 2 /x2 + h
∂x
h′ (x) = 0 =⇒ h(x) = 0
y2
=C
General Solution: xy −
x
(2)2
Apply y(1) = 2 :(1)(2) −
= C =⇒ 2 − 4 = C =⇒ C
1
y2
Solution: xy −
= −2 =⇒ x2 y − y 2 = −2x
x
16