454 Higher Engineering Mathematics
47.4 Further worked problems on
homogeneous first order
differential equations
Problem 3. Solve the equation:
7x(x − y)d y = 2(x 2 + 6x y − 5y 2 )dx
given that x = 1 when y = 0.
Using the procedure of section 47.2:
(i) Rearranging gives:
d y
dx
=
2x
2
+ 12x y − 10y
2
7x 2 − 7x y
which is homogeneous in x and y since each of
the terms on the right hand side is of degree 2.
(ii) Let y = vx then
d y
dx
= v + x
dv
dx
(iii) Substituting for y and
d y
dx
gives:
v + x
dv
dx
=
2x
2
+ 12x(vx) − 10 (vx)
2
7x 2 − 7x(vx)
=
2 + 12v − 10v 2
7 − 7v
(iv) Separating the variables gives:
x
dv
dx
=
2 + 12v − 10v
2
7 − 7v
− v
=
(2 + 12v − 10v 2 ) − v(7 − 7v)
7 − 7v
=
2 + 5v − 3v 2
7 − 7v
Hence,
7 − 7v
2 + 5v − 3v 2 dv =
dx
x
Integrating both sides gives:
7 − 7v
2 + 5v − 3v 2
dv =
1
x
dx
Resolving
7 − 7v
2 + 5v − 3v 2 into partial fractions
gives:
4
(1 + 3v)
−
1
(2 − v)
(see chapter 2)
Hence,
4
(1 + 3v)
−
1
(2 − v)
dv =
1
x
dx
i.e.
4
3
ln(1 + 3v) + ln(2 − v) = ln x + c
(v) Replacing v by
y
x
gives:
4
3
ln
1 +
3y
x
+ ln
2 −
y
x
= ln + c
or
4
3
ln
x + 3y
x
+ ln
2x − y
x
= ln + c
When x = 1, y = 0, thus:
4
3
ln 1 + ln 2 = ln 1 + c
from which, c = ln 2
Hence, the particular solution is:
4
3
ln
x + 3y
x
+ ln
2x − y
x
= ln + ln 2
i.e. ln
x + 3y
x
4
3
2x − y
x
= ln(2x)
from the laws of logarithms
i.e.
x +3y
x
4
3
2x − y
x
= 2x
Problem 4. Show that the solution of the
differential equation: x 2 − 3y 2 + 2x y
d y
dx
= 0 is:
y = x
√
(8x + 1), given that y = 3 when x = 1.
Using the procedure of section 47.2:
(i) Rearranging gives:
2x y
d y
dx
= 3y
2
− x
2 and
d y
dx
=
3y 2 − x 2
2x y
(ii) Let y = vx then
d y
dx
= v + x
dv
dx
(iii) Substituting for y and
d y
dx
gives:
v + x
dv
dx
=
3 (vx)
2
− x 2
2x(vx)
=
3v 2 − 1
2v
(iv) Separating the variables gives:
x
dv
dx
=
3v 2 − 1
2v
− v =
3v 2 − 1 − 2v 2
2v
=
v 2 − 1
2v
Hence,
2v
v 2 − 1
dv =
1
x
dx
Integrating both sides gives:
2v
v 2 − 1
dv =
1
x
dx
i.e. ln(v 2 − 1) = ln x + c
47.4 Further worked problems on
homogeneous first order
differential equations
Problem 3. Solve the equation:
7x(x − y)d y = 2(x 2 + 6x y − 5y 2 )dx
given that x = 1 when y = 0.
Using the procedure of section 47.2:
(i) Rearranging gives:
d y
dx
=
2x
2
+ 12x y − 10y
2
7x 2 − 7x y
which is homogeneous in x and y since each of
the terms on the right hand side is of degree 2.
(ii) Let y = vx then
d y
dx
= v + x
dv
dx
(iii) Substituting for y and
d y
dx
gives:
v + x
dv
dx
=
2x
2
+ 12x(vx) − 10 (vx)
2
7x 2 − 7x(vx)
=
2 + 12v − 10v 2
7 − 7v
(iv) Separating the variables gives:
x
dv
dx
=
2 + 12v − 10v
2
7 − 7v
− v
=
(2 + 12v − 10v 2 ) − v(7 − 7v)
7 − 7v
=
2 + 5v − 3v 2
7 − 7v
Hence,
7 − 7v
2 + 5v − 3v 2 dv =
dx
x
Integrating both sides gives:
7 − 7v
2 + 5v − 3v 2
dv =
1
x
dx
Resolving
7 − 7v
2 + 5v − 3v 2 into partial fractions
gives:
4
(1 + 3v)
−
1
(2 − v)
(see chapter 2)
Hence,
4
(1 + 3v)
−
1
(2 − v)
dv =
1
x
dx
i.e.
4
3
ln(1 + 3v) + ln(2 − v) = ln x + c
(v) Replacing v by
y
x
gives:
4
3
ln
1 +
3y
x
+ ln
2 −
y
x
= ln + c
or
4
3
ln
x + 3y
x
+ ln
2x − y
x
= ln + c
When x = 1, y = 0, thus:
4
3
ln 1 + ln 2 = ln 1 + c
from which, c = ln 2
Hence, the particular solution is:
4
3
ln
x + 3y
x
+ ln
2x − y
x
= ln + ln 2
i.e. ln
x + 3y
x
4
3
2x − y
x
= ln(2x)
from the laws of logarithms
i.e.
x +3y
x
4
3
2x − y
x
= 2x
Problem 4. Show that the solution of the
differential equation: x 2 − 3y 2 + 2x y
d y
dx
= 0 is:
y = x
√
(8x + 1), given that y = 3 when x = 1.
Using the procedure of section 47.2:
(i) Rearranging gives:
2x y
d y
dx
= 3y
2
− x
2 and
d y
dx
=
3y 2 − x 2
2x y
(ii) Let y = vx then
d y
dx
= v + x
dv
dx
(iii) Substituting for y and
d y
dx
gives:
v + x
dv
dx
=
3 (vx)
2
− x 2
2x(vx)
=
3v 2 − 1
2v
(iv) Separating the variables gives:
x
dv
dx
=
3v 2 − 1
2v
− v =
3v 2 − 1 − 2v 2
2v
=
v 2 − 1
2v
Hence,
2v
v 2 − 1
dv =
1
x
dx
Integrating both sides gives:
2v
v 2 − 1
dv =
1
x
dx
i.e. ln(v 2 − 1) = ln x + c