Homogeneous first order differential equations 453
(iv) Separating the variables gives:
x
dv
dx
= v − 1 − v = −1, i.e. dv = −
1
x
dx
Integrating both sides gives:
dv =
−
1
x
dx
Hence, v = −ln x + c
(v) Replacing v by
y
x
gives:
y
x
=−ln x + c, which is
the general solution.
When x = 1, y = 2, thus:
2
1
=−ln 1 + c from
which, c = 2
Thus, the particular solution is:
y
x
=−ln x + 2
or y =−x(ln x −2) or y = x(2 − ln x)
Problem 2. Find the particular solution of the
equation: x
d y
dx
=
x 2 + y 2
y
, given the boundary
conditions that y = 4 when x = 1.
Using the procedure of section 47.2:
(i) Rearranging x
d y
dx
=
x 2 + y 2
y
gives:
d y
dx
=
x 2 + y 2
x y
which is homogeneous in x and y
since each of the three terms on the right hand
side are of the same degree (i.e. degree 2).
(ii) Let y = vx then
d y
dx
= v + x
dv
dx
(iii) Substituting for y and
d y
dx
in the equation
d y
dx
=
x 2 + y 2
x y
gives:
v + x
dv
dx
=
x 2 + v 2 x 2
x(vx)
=
x 2 + v 2 x 2
vx 2
=
1 + v 2
v
(iv) Separating the variables gives:
x
dv
dx
=
1 + v 2
v
− v =
1 + v 2 − v 2
v
=
1
v
Hence, v dv =
1
x
dx
Integrating both sides gives:
v dv =
1
x
dx i.e.
v 2
2
= ln x + c
(v) Replacing v by
y
x
gives:
y 2
2x 2 = ln x + c, which is
the general solution.
When x = 1, y = 4, thus:
16
2
= ln 1 + c from
which, c = 8
Hence, the particular solution is:
y
2
2x 2 = ln x + 8
or y 2 = 2x 2 (8 + lnx)
Now try the following exercise
Exercise 180 Further problems on
homogeneous first order differential
equations
1. Find the general solution of: x 2 = y 2 d y
dx
.
−
1
3
ln
x 3 − y 3
x 3
= ln x + c
2. Find the general solution of:
x − y + x
d y
dx
= 0.
[y = x(c − ln x)]
3. Find the particular solution of the differential equation: (x 2 + y 2 )d y = x y dx, given that
x = 1 when y = 1.
x 2 = 2y 2
ln y +
1
2
4. Solve the differential equation:
x + y
y − x
=
d y
dx
.
⎡
⎣ −
1
2
ln
1 +
2y
x
−
y
2
x 2
= ln x + c
or x 2 + 2x y − y 2 = k
⎤
⎦
5. Find the particular solution of the differential
equation:
2y − x
y + 2x
d y
dx
= 1 given that y = 3
when x = 2.
[x 2 + x y − y 2 = 1]
(iv) Separating the variables gives:
x
dv
dx
= v − 1 − v = −1, i.e. dv = −
1
x
dx
Integrating both sides gives:
dv =
−
1
x
dx
Hence, v = −ln x + c
(v) Replacing v by
y
x
gives:
y
x
=−ln x + c, which is
the general solution.
When x = 1, y = 2, thus:
2
1
=−ln 1 + c from
which, c = 2
Thus, the particular solution is:
y
x
=−ln x + 2
or y =−x(ln x −2) or y = x(2 − ln x)
Problem 2. Find the particular solution of the
equation: x
d y
dx
=
x 2 + y 2
y
, given the boundary
conditions that y = 4 when x = 1.
Using the procedure of section 47.2:
(i) Rearranging x
d y
dx
=
x 2 + y 2
y
gives:
d y
dx
=
x 2 + y 2
x y
which is homogeneous in x and y
since each of the three terms on the right hand
side are of the same degree (i.e. degree 2).
(ii) Let y = vx then
d y
dx
= v + x
dv
dx
(iii) Substituting for y and
d y
dx
in the equation
d y
dx
=
x 2 + y 2
x y
gives:
v + x
dv
dx
=
x 2 + v 2 x 2
x(vx)
=
x 2 + v 2 x 2
vx 2
=
1 + v 2
v
(iv) Separating the variables gives:
x
dv
dx
=
1 + v 2
v
− v =
1 + v 2 − v 2
v
=
1
v
Hence, v dv =
1
x
dx
Integrating both sides gives:
v dv =
1
x
dx i.e.
v 2
2
= ln x + c
(v) Replacing v by
y
x
gives:
y 2
2x 2 = ln x + c, which is
the general solution.
When x = 1, y = 4, thus:
16
2
= ln 1 + c from
which, c = 8
Hence, the particular solution is:
y
2
2x 2 = ln x + 8
or y 2 = 2x 2 (8 + lnx)
Now try the following exercise
Exercise 180 Further problems on
homogeneous first order differential
equations
1. Find the general solution of: x 2 = y 2 d y
dx
.
−
1
3
ln
x 3 − y 3
x 3
= ln x + c
2. Find the general solution of:
x − y + x
d y
dx
= 0.
[y = x(c − ln x)]
3. Find the particular solution of the differential equation: (x 2 + y 2 )d y = x y dx, given that
x = 1 when y = 1.
x 2 = 2y 2
ln y +
1
2
4. Solve the differential equation:
x + y
y − x
=
d y
dx
.
⎡
⎣ −
1
2
ln
1 +
2y
x
−
y
2
x 2
= ln x + c
or x 2 + 2x y − y 2 = k
⎤
⎦
5. Find the particular solution of the differential
equation:
2y − x
y + 2x
d y
dx
= 1 given that y = 3
when x = 2.
[x 2 + x y − y 2 = 1]