Higher Engineering Mathematics, Sixth Edition

454 Higher Engineering Mathematics

47.4 Further worked problems on

homogeneous first order

differential equations

Problem 3. Solve the equation:

7 x(x−y)dy= 2 (x^2 + 6 xy− 5 y^2 )dx

given thatx=1wheny=0.

Using the procedure of section 47.2:

(i) Rearranging gives:

dy

dx

=

2 x^2 + 12 xy− 10 y^2

7 x^2 − 7 xy

which is homogeneous inxandysince each of

the terms on the right hand side is of degree 2.

(ii) Lety=vxthen

dy

dx

=v+x

dv

dx

(iii) Substituting foryand

dy

dx

gives:

v+x

dv

dx

=

2 x^2 + 12 x(vx)− 10 (vx)^2

7 x^2 − 7 x(vx)

=

2 + 12 v− 10 v^2

7 − 7 v

(iv) Separating the variables gives:

x

dv

dx

=

2 + 12 v− 10 v^2

7 − 7 v

−v

=

( 2 + 12 v− 10 v^2 )−v( 7 − 7 v)

7 − 7 v

=

2 + 5 v− 3 v^2

7 − 7 v

Hence,

7 − 7 v

2 + 5 v− 3 v^2

dv=

dx

x

Integrating both sides gives:

∫ (

7 − 7 v

2 + 5 v− 3 v^2

)

dv=

∫

1

x

dx

Resolving

7 − 7 v

2 + 5 v− 3 v^2

into partial fractions

gives:

4

( 1 + 3 v)

−

1

( 2 −v)

(see chapter 2)

Hence,

∫ (

4

( 1 + 3 v)

−

1

( 2 −v)

)

dv=

∫

1

x

dx

i.e.

4

3

ln( 1 + 3 v)+ln( 2 −v)=lnx+c

(v) Replacingvby

y

x

gives:

4

3

ln

(

1 +

3 y

x

)

+ln

(

2 −

y

x

)

=ln+c

or

4

3

ln

(

x+ 3 y

x

)

+ln

(

2 x−y

x

)

=ln+c

Whenx= 1 ,y=0, thus:

4

3

ln1+ln2=ln1+c

from which,c=ln 2

Hence, the particular solution is:

4

3

ln

(

x+ 3 y

x

)

+ln

(

2 x−y

x

)

=ln+ln2

i.e. ln

(

x+ 3 y

x

)^43 (

2 x−y

x

)

=ln( 2 x)

from the laws of logarithms

i.e.

(

x+ 3 y

x

) 43 (

2 x−y

x

)

= 2 x

Problem 4. Show that the solution of the

differential equation:x^2 − 3 y^2 + 2 xy

dy

dx

=0is:

y=x

√

( 8 x+ 1 ), given thaty=3whenx=1.

Using the procedure of section 47.2:

(i) Rearranging gives:

2 xy

dy

dx

= 3 y^2 −x^2 and

dy

dx

=

3 y^2 −x^2

2 xy

(ii) Lety=vxthen

dy

dx

=v+x

dv

dx

(iii) Substituting foryand

dy

dx

gives:

v+x

dv

dx

=

3 (vx)^2 −x^2

2 x(vx)

=

3 v^2 − 1

2 v

(iv) Separating the variables gives:

x

dv

dx

=

3 v^2 − 1

2 v

−v=

3 v^2 − 1 − 2 v^2

2 v

=

v^2 − 1

2 v

Hence,

2 v

v^2 − 1

dv=

1

x

dx

Integrating both sides gives:

∫

2 v

v^2 − 1

dv=

∫

1

x

dx

i.e. ln(v^2 − 1 )=lnx+c