Higher Engineering Mathematics, Sixth Edition

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

1. Solve the differential equation:
   xy^3 dy=(x^4 +y^4 )dx. [
   y^4 = 4 x^4 (lnx+c)

]

2. 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

]

3. Solve the differential equation:
   2 x

dy

dx

=x+ 3 y, given that whenx=1,y=1.

[

(x+y)^2 = 4 x^3

]

4. 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.

5. Determine the particular solution of
   dy
   dx

=

x^3 +y^3

xy^2

, given thatx=1wheny=4.

[

y^3 =x^3 (3lnx+ 64 )

]

6. 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.