Homogeneous first order differential equations 455
(v) Replacing v by
y
x
gives:
ln
y
2
x 2 − 1
= ln x + c,
which is the general solution.
When y = 3, x = 1, thus: ln
9
1
− 1
= ln 1 +c
from which, c = ln 8
Hence, the particular solution is:
ln
y 2
x 2 − 1
= ln x + ln 8 = ln 8x
by the laws of logarithms
Hence,
y 2
x 2 − 1
= 8x i.e.
y 2
x 2 = 8x + 1 and
y 2 = x 2 (8x + 1)
i.e.
y = x
√
(8x + 1)
Now try the following exercise
Exercise 181 Further problems on
homogeneous first order differential
equations
1. Solve the differential equation:
x y 3 d y = (x 4 + y 4 )dx.
y 4 = 4x 4 (ln x + c)
2. Solve: (9x y − 11x y)
d y
dx
= 11y
2
− 16x y + 3x
2 .
1
5
3
13
ln
13y − 3x
x
− ln
y − x
x
= ln x + c
3. Solve the differential equation:
2x
d y
dx
= x + 3y, given that when x = 1, y = 1.
(x + y)
2
= 4x
3
4. Show that the solution of the differential equation: 2x y
d y
dx
= x
2
+ y
2 can be expressed as:
x = K(x 2 − y 2 ), where K is a constant.
5. Determine the particular solution of
d y
dx
=
x 3 + y 3
x y 2 , given that x = 1 when y = 4.
y
3
= x
3
(3 ln x + 64)
6. Show that the solution of the differential equation:
d y
dx
=
y 3 − x y 2 − x 2 y − 5x 3
x y 2 − x 2 y − 2x 3
is of the
form:
y 2
2x 2 +
4y
x
+ 18 ln
y − 5x
x
= ln x + 42,
when x = 1 and y = 6.