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