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 6= 0
x−y
ax − ay
=
.
ax
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
dv
=
x
1−v
By integrating we obtain
log |x| = − log |1 − v|,
or
x(1 − v) = c.
Finally,
x − y = c.
Exercise 4.3.
(x2 − y 2 )dx + 2xydy = 0.
Solution. We have a homogeneous equations. Integrating factor for this equation is
1
1
=
.
I(x, y) = 3
x − xy 2 + 2xy 2
x(x2 + y 2 )
1
2
Thus, we have the exact equation
x2 − y 2
2y
dx + 2
dy = 0.
x(x2 + y 2 )
x + y2
Taking the initial point (x0 , y0 ), x0 > 0, y0 = 0, by integrating we obtain
Z x
Z y
du
2v
+
dv = c1 ,
2
2
x0 u
0 x +v
or
log x − log x0 + log(x2 + y 2 ) − log x2 = c1 ,
log(x2 + y 2 ) = c2 + log x,
x2 + y 2 = cx.
Exercise 4.4.
p
x2 + y 2 dx = xdy − ydx.
Solution. This is a homogeneous equation. Let us find an integrating factor.
1
1
I(x, y) = p
= p
.
2
2
2
x x + y + xy − yx
x x + y2
Therefore, the differential equation
dx
ydx
dy
+ p
−p
= 0.
2
2
x
x x +y
x2 + y 2
By integrating we obtain
Z
x
du
−
u
Z
y
dv
= c1 ,
+ v2
x0
0
p
log |x| − log |x0 | − log |y + x2 + y 2 | − log |x| = c1
Finally,
y+
√
x2
p
x2 + y 2 = c.
Exercise 4.5.
(x2 y + 2xy 2 − y 3 )dx − (2y 3 − xy 2 + x3 )dy = 0.
Solution. The differential equation is homogeneous. Denote y = vx. Then
(x3 v + 2x3 v 2 − x3 v 3 )dx − (2x3 v 3 − x3 v 2 + x3 )(vdx + xdv) = 0,
(x3 v + 2x3 v 2 − x3 v 3 )dx − (2x3 v 4 − x3 v 3 + x3 v)dx − (2x4 v 3 − x4 v 2 + x4 )dv = 0,
x3 (2v 2 − 2v 4 )dx − x4 (2v 3 − v 2 + 1)dv = 0,
2v 3 − v 2 + 1
dx
=
dv.
x
2v 2 − 2v 4
1
2 log |x| = c1 − − log |1 − v 2 |,
v
x2 e−v (1 − v 2 ) = c,
c = (x2 − y 2 )ex/y .
Exercise 4.6.
³
y
y´
y
x sin − y cos
dx + x cos dy = 0.
x
x
x
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
= − tan vdv.
x
By integrating,
log c|x| = log | cos v|,
cx = cos v,
y
cx = cos .
x
Exercise 4.7.
(x3 + 2xy 2 )dx + (y 3 + 2x2 y)dy = 0.
Solution. It is readily seen that the differential equation is homogeneous. The
integrating factor is
1
1
I(x, y) = 4
= 4
.
2
2
4
2
2
x + 2x y + y + 2x y
x + 4x2 y 2 + y 4
Therefore, the differential equation
x3 + 2xy 2
y 3 + 2x2 y
dx
+
dy = 0
x4 + 4x2 y 2 + y 4
x4 + 4x2 y 2 + y 4
is exact. Its integrating yields
Z x
Z y
du
v 3 + 2x2 v
+
dv = c1 .
4
2 2
4
u
0
0 x + 4x v + v
The last expression can be rewritten as follows
Z x
Z
du 1 y d(v 4 + 4x2 v 2 + x4 )
+
= c1 ,
u
4 0 x4 + 4x2 v 2 + v 4
0
or finally
cx = x4 + 4x2 y 2 + y 4 .
Exercise 4.8.
(4x4 − x3 y + y 4 )dx + x4 dy = 0.
Solution. It is readily seen that the differential equation is homogeneous. Denote y = vx. Then,
(4x4 − x4 v + x4 v 4 )dx + x4 (xdv + vdx) = 0,
or
(4x4 + x4 v 4 )dx + x5 dv = 0,
dv
dx
=−
,
x
4Z+ v 4
dv
log |cx| = −
,
4 + v4
Z
³
dv ´
cx = exp −
.
4 + v4
Exercise 4.9.
³
x2 sin
y2
y2 ´
y2
2
−
2y
cos
dx + 2xy cos 2 dy = 0.
2
2
x
x
x
4
Solution. The differential equation is homogeneous. Denote y = xv. Then,
(x2 sin v 2 − 2x2 v 2 cos v 2 )dx + 2x2 v cos v 2 (xdv + vdx) = 0,
or
sin v 2 dx + 2v cos v 2 (xdv) = 0,
dx
= −2v cot v 2 dv,
x
Z
d(sin v 2 )
log |x| = −
= − log | sin v 2 | + log c,
sin v 2
y2
x sin 2 = c.
x
Exercise 4.10.
2
2
(x2 e−y /x − y 2 )dx + xydy = 0.
Solution. The differential equation is homogeneous. Then,
2
I(x, y) =
x3 e−y2 /x2
1
ey /x
=
2
2
x3
− y x + xy
2
is an integrating factor. Next,
2
2
³ 1 y 2 ey2 /x2 ´
yey /x
−
dx +
dy = 0.
x
x3
x2
is an exact differential equation. Therefore,
Z x
Z y
2
2
1
du
+ 2
vev /x dv = c1 ,
x 0
x0 6=0 u
1 2 2 1
log x − log x0 + ev /x − = c1 ,
2
2
³
y2 ´
cx = exp exp 2 .
x
Exercise 4.11.
(2x + y − 2)dx + (2y − x + 1)dy = 0.
Solution. The differential equation is not homogeneous. To reduce it to homogeneous, 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
1
1
= 2
.
I(u, v) = 2
2
2u + uv + 2v − uv
2u + 2v 2
Then,
2u + v
2v − u
du + 2
dv = 0
2
2
2u + 2v
2u + 2v 2
5
is an exact equation, and
Z
Z v
dt
2z − u
+
dz
2 + 2z 2
t
2u
u0 6=0
0
Z
Z v
´
1 ³ v d(2z 2 + 2u2 )
2u
= log |u| − log |u0 | +
dz
−
dz
2
2
2
2
2 0 2u + 2z
0 2u + 2z
1³
v´
= log u − log u0 +
log(2z 2 + 2u2 ) − log(2u2 ) − arctan
= c1 ,
2
u
or finally
y
= c.
log[(x − 1)2 + y 2 ] − arctan
x−1
Exercise 4.12.
(x − 3y)dx + (x + y − 4)dy = 0.
Solution. The differential equation is not homogeneous. To reduce it to homogeneous, let us put x = u + h, y = v + k. Then,
u
(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
1
1
I(x, y) = 2
=
.
u − 3uv + uv + v 2
(u − v)2
Thus, we have the following exact equation
u − 3v
u+v
du +
dv = 0.
(u − v)2
(u − v)2
By integrating we obtain
Z
v
log |u| − log |u0 | +
Z
0
t+u
dt
(t − u)2
Z v
t−u
d(t − u)
= log |u| − log |u0 | +
d(t − u) + 2u
2
2
0 (t − u)
0 (t − u)
2u
= log |u| − log |u0 | + log |v − u| − log |u| −
+ 2 = c1 .
v−u
v
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
or
u
du = −dy.
2(u + 1)
By integrating we obtain
Z
y − c1 = −
Finally,
Z
Z
u
1
u+1
du =
du −
du
2(u + 1)
2(u + 1)
2(u + 1)
1
1
= log |u + 1| − (u + 1).
2
2
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,
u2
y+u−
= c/2,
2
or
2x + 6y − (x + 2y)2 = c.
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