Homogeneous first order differential equations 455
(v) Replacing v by
y
x
gives:
ln
(
y^2
x^2
− 1
)
=lnx+c,
which is the general solution.
When y= 3 ,x=1, thus: ln
(
9
1
− 1
)
=ln1+c
from which,c=ln 8
Hence, the particular solution is:
ln
(
y^2
x^2
− 1
)
=lnx+ln8=ln8x
by the laws of logarithms
Hence,
(
y^2
x^2
− 1
)
= 8 x i.e.
y^2
x^2
= 8 x+1and
y^2 =x^2 ( 8 x+ 1 )
i.e. y=x
√
( 8 x+ 1 )
Now try the following exercise
Exercise 181 Further problems on
homogeneous first order differential
equations
- Solve the differential equation:
xy^3 dy=(x^4 +y^4 )dx. [
y^4 = 4 x^4 (lnx+c)
]
- Solve:( 9 xy− 11 xy)
dy
dx
= 11 y^2 − 16 xy+ 3 x^2.
[
1
5
{
3
13
ln
(
13 y− 3 x
x
)
−ln
(
y−x
x
)}
=lnx+c
]
- Solve the differential equation:
2 x
dy
dx
=x+ 3 y, given that whenx=1,y=1.
[
(x+y)^2 = 4 x^3
]
- Show that the solutionof the differential equa-
tion: 2xy
dy
dx
=x^2 +y^2 can be expressed as:
x=K(x^2 −y^2 ),where K is a constant.
- Determine the particular solution of
dy
dx
=
x^3 +y^3
xy^2
, given thatx=1wheny=4.
[
y^3 =x^3 (3lnx+ 64 )
]
- Show that the solutionof the differential equa-
tion:
dy
dx
=
y^3 −xy^2 −x^2 y− 5 x^3
xy^2 −x^2 y− 2 x^3
is of the
form:
y^2
2 x^2
+
4 y
x
+18ln
(
y− 5 x
x
)
=lnx+ 42 ,
whenx=1andy=6.