differential equation

454 DIFFERENTIAL EQUATIONS
( ) 9
When y = 3, x = 1, thus: ln
1 − 1 = ln 1 + c
from which, c = ln 8
Hence, the particular solution is:
( y
2
)
ln
x 2 − 1 = ln x + ln 8 = ln 8x
by the laws of logarithms.
( y
2
)
Hence,
x 2 − 1 y 2
= 8x i.e. = 8x + 1 and
x2 y 2 = x 2 (8x + 1)
i.e. y = x √ (8x + 1)
Now try the following exercise.
Exercise 182 Further problems on homogeneous
first order differential equations
1. Solve the differential equation:
xy 3 dy = (x 4 + y 4 )dx [
y 4 = 4x 4 (ln x + c) ]
3. Solve the differential equation:
2x dy = x + 3y, given that when x = 1, y = 1.
dx
[ (x + y) 2 = 4x 3]
4. Show that the solution of the differential
equation: 2xy dy
dx = x2 + y 2 can be expressed
as: x = K(x 2 − y 2 ), where K is a constant.
5. Determine the particular solution of
dy
dx = x3 + y 3
xy 2 , given that x = 1 when y = 4.
[
y 3 = x 3 (3 ln x + 64) ]
6. Show that the solution of the differential
dy
equation:
dx = y3 − xy 2 − x 2 y − 5x 3
xy 2 − x 2 y − 2x 3 is of
the form:
y 2
2x 2 + 4y ( ) y − 5x
x
+ 18 ln = ln x + 42,
x
when x = 1 and y = 6.
2. Solve: (9xy − 11xy) dy
dx = 11y2 − 16xy + 3x 2
[ { ( ) ( )}
1 3 13y − 3x y − x
5 13 ln − ln
x
x
]
= ln x + c