Higher Engineering Mathematics, Sixth Edition

Homogeneous first order differential equations 453

(iv) Separating the variables gives:

x

dv

dx

=v− 1 −v=− 1 , i.e.dv=−

1

x

dx

Integrating both sides gives:

∫

dv=

∫

−

1

x

dx

Hence,v=−lnx+c

(v) Replacingvby

y

x

gives:

y

x

=−lnx+c,whichis

the general solution.

When x= 1 ,y=2, thus:

2

1

=−ln1+c from

which,c= 2

Thus, the particular solution is:

y

x

=−lnx+ 2

ory=−x(lnx−2)ory=x( 2 −lnx)

Problem 2. Find the particular solution of the

equation:x

dy

dx

=

x^2 +y^2

y

, given the boundary

conditions thaty=4whenx=1.

Using the procedure of section 47.2:

(i) Rearrangingx

dy

dx

=

x^2 +y^2

y

gives:

dy

dx

=

x^2 +y^2

xy

which is homogeneous inxandy

since each of the three terms on the right hand

side are of the same degree (i.e. degree 2).

(ii) Lety=vxthen

dy

dx

=v+x

dv

dx

(iii) Substituting for y and

dy

dx

in the equation

dy

dx

=

x^2 +y^2

xy

gives:

v+x

dv

dx

=

x^2 +v^2 x^2

x(vx)

=

x^2 +v^2 x^2

vx^2

=

1 +v^2

v

(iv) Separating the variables gives:

x

dv

dx

=

1 +v^2

v

−v=

1 +v^2 −v^2

v

=

1

v

Hence,vdv=

1

x

dx

Integrating both sides gives:

∫

vdv=

∫

1

x

dxi.e.

v^2

2

=lnx+c

(v) Replacingvby

y

x

gives:

y^2

2 x^2

=lnx+c,whichis

the general solution.

When x= 1 ,y=4, thus:

16

2

=ln1+c from

which,c= 8

Hence, the particular solution is:

y^2

2 x^2

=lnx+ 8

ory^2 =2x^2 ( 8 +lnx)

Now try the following exercise

Exercise 180 Further problems on

homogeneous first order differential

equations

1. Find the general solution of: x^2 =y^2

dy

dx

.

[

−

1

3

ln

(

x^3 −y^3

x^3

)

=lnx+c

]

2. Find the general solution of:
   x−y+x

dy

dx

= 0. [y=x(c−lnx)]

3. Find the particular solution of the differen-
   tial equation:(x^2 +y^2 )dy=xydx, given that
   x=1wheny=1.
   [
   x^2 = 2 y^2

(

lny+

1

2

)]

4. Solve the differential equation:

x+y

y−x

=

dy

dx

.

⎡

⎣−

1

2

ln

(

1 +

2 y

x

−

y^2

x^2

)

=lnx+c

orx^2 + 2 xy−y^2 =k

⎤

⎦

5. Find the particular solution of the differential
   equation:

(

2 y−x

y+ 2 x

)

dy

dx

=1 given thaty= 3

whenx=2. [x^2 +xy−y^2 =1]