solucionario

1. Lesson 4: Homogeneous differential equations of the first order
Solve the following differential equations
Exercise 4.1.
(x − y)dx + xdy = 0.
Solution.
The coefficients of the differential equations are homogeneous, since for any a = 0
ax − ay
ax
=
x − y
x
.
Then denoting y = vx we obtain
(1 − v)xdx + vxdx + x2
dv = 0,
or
xdx + x2
dv = 0.
By integrating we have
x = e−v
+ c,
or finally
x = e−y/x
+ c
Exercise 4.2.
(x − 2y)dx + xdy = 0.
Solution.
It is easily seen that the differential equation is homogeneous. Then denoting
y = vx we obtain
(x − 2xv)dx + xvdx + x2
dv = 0.
That is
x(1 − v)dx + x2
dv = 0.
It is easily seen that an integrating factor is
1
x2(1 − v)
.
Therefore,
dx
x
=
dv
1 − v
By integrating we obtain
log |x| = − log |1 − v|,
or
x(1 − v) = c.
Finally,
x − y = c.
Exercise 4.3.
(x2
− y2
)dx + 2xydy = 0.
Solution. We have a homogeneous equations. Integrating factor for this equa-
tion is
I(x, y) =
1
x3 − xy2 + 2xy2
=
1
x(x2 + y2)
.
1

2. 2
Thus, we have the exact equation
x2
− y2
x(x2 + y2)
dx +
2y
x2 + y2
dy = 0.
Taking the initial point (x0, y0), x0 > 0, y0 = 0, by integrating we obtain
x
x0
du
u
+
y
0
2v
x2 + v2
dv = c1,
or
log x − log x0 + log(x2
+ y2
) − log x2
= c1,
log(x2
+ y2
) = c2 + log x,
x2
+ y2
= cx.
Exercise 4.4.
x2 + y2dx = xdy − ydx.
Solution. This is a homogeneous equation. Let us find an integrating factor.
I(x, y) =
1
x x2 + y2 + xy − yx
=
1
x x2 + y2
.
Therefore, the differential equation
dx
x
+
ydx
x x2 + y2
−
dy
x2 + y2
= 0.
By integrating we obtain
x
x0
du
u
−
y
0
dv
√
x2 + v2
= c1,
log |x| − log |x0| − log |y + x2 + y2| − log |x| = c1
Finally,
y + x2 + y2 = c.
Exercise 4.5.
(x2
y + 2xy2
− y3
)dx − (2y3
− xy2
+ x3
)dy = 0.
Solution. The differential equation is homogeneous. Denote y = vx. Then
(x3
v + 2x3
v2
− x3
v3
)dx − (2x3
v3
− x3
v2
+ x3
)(vdx + xdv) = 0,
(x3
v + 2x3
v2
− x3
v3
)dx − (2x3
v4
− x3
v3
+ x3
v)dx − (2x4
v3
− x4
v2
+ x4
)dv = 0,
x3
(2v2
− 2v4
)dx − x4
(2v3
− v2
+ 1)dv = 0,
dx
x
=
2v3
− v2
+ 1
2v2 − 2v4
dv.
2 log |x| = c1 −
1
v
− log |1 − v2
|,
x2
e−v
(1 − v2
) = c,
c = (x2
− y2
)ex/y
.
Exercise 4.6.
x sin
y
x
− y cos
y
x
dx + x cos
y
x
dy = 0.

3. 3
Solution. It is readily seen that the differential equation is homogeneous.
Putting y = xv we obtain
(x sin v − xv cos v)dx + x cos v(xdv + vdx) = 0,
sin vdx + x cos vdv = 0,
dx
x
= − tan vdv.
By integrating,
log c|x| = log | cos v|,
cx = cos v,
cx = cos
y
x
.
Exercise 4.7.
(x3
+ 2xy2
)dx + (y3
+ 2x2
y)dy = 0.
Solution. It is readily seen that the differential equation is homogeneous. The
integrating factor is
I(x, y) =
1
x4 + 2x2y2 + y4 + 2x2y2
=
1
x4 + 4x2y2 + y4
.
Therefore, the differential equation
x3
+ 2xy2
x4 + 4x2y2 + y4
dx +
y3
+ 2x2
y
x4 + 4x2y2 + y4
dy = 0
is exact. Its integrating yields
x
0
du
u
+
y
0
v3
+ 2x2
v
x4 + 4x2v2 + v4
dv = c1.
The last expression can be rewritten as follows
x
0
du
u
+
1
4
y
0
d(v4
+ 4x2
v2
+ x4
)
x4 + 4x2v2 + v4
= c1,
or finally
cx = x4
+ 4x2
y2
+ y4
.
Exercise 4.8.
(4x4
− x3
y + y4
)dx + x4
dy = 0.
Solution. It is readily seen that the differential equation is homogeneous. De-
note y = vx. Then,
(4x4
− x4
v + x4
v4
)dx + x4
(xdv + vdx) = 0,
or
(4x4
+ x4
v4
)dx + x5
dv = 0,
dx
x
= −
dv
4 + v4
,
log |cx| = −
dv
4 + v4
,
cx = exp −
dv
4 + v4
.
Exercise 4.9.
x2
sin
y2
x2
− 2y2
cos
y2
x2
dx + 2xy cos
y2
x2
dy = 0.

4. 4
Solution. The differential equation is homogeneous. Denote y = xv. Then,
(x2
sin v2
− 2x2
v2
cos v2
)dx + 2x2
v cos v2
(xdv + vdx) = 0,
or
sin v2
dx + 2v cos v2
(xdv) = 0,
dx
x
= −2v cot v2
dv,
log |x| = −
d(sin v2
)
sin v2
= − log | sin v2
| + log c,
x sin
y2
x2
= c.
Exercise 4.10.
(x2
e−y2
/x2
− y2
)dx + xydy = 0.
Solution. The differential equation is homogeneous. Then,
I(x, y) =
1
x3e−y2/x2
− y2x + xy2
=
ey2
/x2
x3
is an integrating factor. Next,
1
x
−
y2
ey2
/x2
x3
dx +
yey2
/x2
x2
dy = 0.
is an exact differential equation. Therefore,
x
x0=0
du
u
+
1
x2
y
0
vev2
/x2
dv = c1,
log x − log x0 +
1
2
ev2
/x2
−
1
2
= c1,
cx = exp exp
y2
x2
.
Exercise 4.11.
(2x + y − 2)dx + (2y − x + 1)dy = 0.
Solution. The differential equation is not homogeneous. To reduce it to homo-
geneous, let us put x = u + h, y = v + k. Then,
(2u + 2h + v + k − 2)dx + (2v + 2k − u − h + 1)dy = 0.
Then we have the following system
2h + k = 2,
2k − h = −1.
We have k = 0, h = 1, and therefore,
(2u + v)du + (2v − u)dv = 0
is a homogeneous differential equation. The integrating factor is
I(u, v) =
1
2u2 + uv + 2v2 − uv
=
1
2u2 + 2v2
.
Then,
2u + v
2u2 + 2v2
du +
2v − u
2u2 + 2v2
dv = 0

5. 5
is an exact equation, and
u
u0=0
dt
t
+
v
0
2z − u
2u2 + 2z2
dz
= log |u| − log |u0| +
1
2
v
0
d(2z2
+ 2u2
)
2u2 + 2z2
dz −
v
0
2u
2u2 + 2z2
dz
= log u − log u0 +
1
2
log(2z2
+ 2u2
) − log(2u2
) − arctan
v
u
= c1,
or finally
log[(x − 1)2
+ y2
] − arctan
y
x − 1
= c.
Exercise 4.12.
(x − 3y)dx + (x + y − 4)dy = 0.
Solution. The differential equation is not homogeneous. To reduce it to homo-
geneous, let us put x = u + h, y = v + k. Then,
(u − 3v + h − 3k)du + (u + h + v + k − 4)dv = 0.
Having a system of equations
h = 3k
h + k = 4
we obtain h = 3 and k = 1. Therefore, the new differential equation
(u − 3v)du + (u + v)dv = 0.
is homogeneous. The integrating factor is
I(x, y) =
1
u2 − 3uv + uv + v2
=
1
(u − v)2
.
Thus, we have the following exact equation
u − 3v
(u − v)2
du +
u + v
(u − v)2
dv = 0.
By integrating we obtain
log |u| − log |u0| +
v
0
t + u
(t − u)2
dt
= log |u| − log |u0| +
v
0
t − u
(t − u)2
d(t − u) + 2u
v
0
d(t − u)
(t − u)2
= log |u| − log |u0| + log |v − u| − log |u| −
2u
v − u
+ 2 = c1.
Therefore,
c2(u − v) log |u − v| = u,
or finally,
c(x − y − 2) log |x − y − 2| = x,
Exercise 4.13.
(x − y)dx + (x − y + 2)dy = 0.
Solution. The differential equation is not homogeneous. Putting x − y = u we
have
udu + 2(u + 1)dy = 0,

6. 6
or
u
2(u + 1)
du = −dy.
By integrating we obtain
y − c1 = −
u
2(u + 1)
du =
1
2(u + 1)
du −
u + 1
2(u + 1)
du
=
1
2
log |u + 1| −
1
2
(u + 1).
Finally,
x + y − log |x + y − 1| = c.
Exercise 4.14.
(x + 2y + 1)dx + (2x + 4y + 3)dy = 0.
Solution. The differential equation is not homogeneous. Putting x+2y = u we
have
(u + 1)du − 2(u + 1)dy + (2u + 1)dy + dy = 0,
or
(u + 1)du + dy = 0.
Therefore,
y + u −
u2
2
= c/2,
or
2x + 6y − (x + 2y)2
= c.
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