Higher Engineering Mathematics

454 DIFFERENTIAL EQUATIONS

Wheny=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

)

=lnx+ln 8=ln 8x

by the laws of logarithms.

Hence,

(

y^2

x^2

− 1

)

= 8 x i.e.

y^2

x^2

= 8 x+1 and

y^2 =x^2 ( 8 x+ 1 )

i.e. y=x

√

( 8 x+ 1 )

Now try the following exercise.

Exercise 182 Further problems on homoge-

neous 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: (9xy− 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 solution of the differential
   equation: 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=1 wheny=4.

[

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

]

6. Show that the solution of the differential

equation:

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

+18 ln

(

y− 5 x

x

)

=lnx+42,

whenx=1 andy=6.