homogeneous Equation All Math Solved,Awsome Solution

1. Differential Equations:
First order and first degree D.E.
(Homogeneous D.E)
8/8/2020 1
Lecture # 6

2. Homogeneous D.E.
8/8/2020 2
Homogeneous D.Es: A differential equation of the form
dy
dx
=
f1
(x, y)
f2
(x, y)
where f1
(x, y) and f2
(x, y) are homogeneous
functions of x and y of same degree is called a homogeneous
D.E.
To solve this kind of D.E., we put
y = vx
i.e.
dy
dx
= v + x
dv
dx
to separate the variables.
For example-
dy
dx
=
x2
+ y2
2xy
is a homogeneous D.E.

3. 8/8/2020 3
Solve the following Homogeneous D.Es:
(a) (x2
+ y2
)dx +2xydy =0 (b) y2
+ x2 dy
dx
= xy
dy
dx
(c)
1
2x
dy
dx
+
x + y
x2
+ y2
=0 (d)
dy
dx
+
y(x + y)
x2
=0
(e) x2
ydx - x3
dy = y3
dy
Homogeneous D.E.

4. Solution (a):
8/8/2020 4
Given that
(x2
+ y2
)dx +2xydy =0
We can write
dy
dx
= -
x2
+ y2
2xy
...........(1) which is a homogeneous D.E.
So putting y = vx
dy
dx
= v + x
dv
dx
Now putting these values in Eq. (1), we get
v + x
dv
dx
= -
x2
+(vx)2
2x ×vx
= -
x2
+ v2
x2
2x2
v
Þ x
dv
dx
= -
1+ v2
2v
- v
Homogeneous D.E.

5. 8/8/2020 5
Þ x
dv
dx
=
-1- v2
-2v2
2v
Þ x
dv
dx
=
-(1+3v2
)
2v
Now integrating, we get
2v
1+3v2
dvò = -
1
x
dxò
Þ
1
3
3×2v
1+3v2
dvò = -
1
x
dxò
Now separating the variables, we get
2v
1+3v2
dv = -
1
x
dx
Homogeneous D.E.
Þ
1
3
log(1+3v2
)= -logx +c
Þlog(1+3v2
)1/3
= logx-1
+log A
Þlog(1+3v2
)1/3
= log(x-1
A)

6. 8/8/2020 6
Þ(1+3v2
)1/3
=
A
x
Þ1+3v2
=
A3
x3
Þ1+3
y2
x2
=
A3
x3
[∵ y = vx]
Homogeneous D.E.
Þ
x2
+3y2
x2
=
A3
x3
Þ x(x2
+3y2
)= A3
(Ans.)

7. Solution (b):
8/8/2020 7
Given that
y2
+ x2 dy
dx
= xy
dy
dx
We can write
dy
dx
=
y2
xy - x2
...........(1) which is a homogeneous D.E.
So putting y = vx
dy
dx
= v + x
dv
dx
Now putting these values in Eq. (1), we get
v + x
dv
dx
=
(vx)2
x ×vx - x2
=
v2
x2
x2
(v -1)
Þ x
dv
dx
=
v2
v -1
Homogeneous D.E.

8. 8/8/2020 8
Now integrating, we get
1
v
-
1
v2
æ
èç
ö
ø÷ dvò =
1
x
dxò
Þlogv +
1
v
= logx +c
Now separating the variables, we get
v -1
v2
dv =
1
x
dx
Homogeneous D.E.
Þlogv +loge1/v
= logx +log A
Þlog(ve1/v
)= log(xA)
Þve1/v
= xA
Þ
1
v
-
1
v2
æ
èç
ö
ø÷dv =
1
x
dx
Þ
y
x
ex/y
= xA [∵ y = vx]

9. Solution (b):
8/8/2020 9
Given that
dy
dx
+
y(x + y)
x2
= 0
We can write
dy
dx
= -
xy + y2
x2
...........(1) which is a homogeneous D.E.
So putting y = vx
dy
dx
= v + x
dv
dx
Now putting these values in Eq. (1), we get
v + x
dv
dx
= -
x ×vx +(vx)2
x2
= -
vx2
+ v2
x2
x2
Þ x
dv
dx
= -v - v2
- v Þ x
dv
dx
= -(v2
+2v)
Homogeneous D.E.

10. 8/8/2020 10
Now integrating, we get
1
(v +1)2
-12
dvò = -
1
x
dxò
Now separating the variables, we get
1
v2
+2v
dv = -
1
x
dx
Þ
1
v2
+2v +1-1
dv = -
1
x
dx
Homogeneous D.E.
Þ
1
(v +1)2
-12
dv = -
1
x
dx

11. 8/8/2020 11
Þ
1
2
log
v +1-1
v +1+1
= -log x +c ∵
1
x2
-a2
dx =
1
2aò log
x -a
x +a
é
ë
ê
ù
û
ú
Þ
1
2
log
v
v +2
= log x-1
+log A
Homogeneous D.E.
Þlog
v
v +2
æ
èç
ö
ø÷
1/2
= log
A
x
Þ
v
v +2
=
A2
x2
Þ
y / x
y / x +2
=
A2
x2
[∵ y = vx]
Þ
y
y +2x
=
A2
x2
Þ yx2
= A2
( y +2x) (Ans.)

12. Exercise the following Problems:
8/8/2020 12
(a) x2
ydx -(x3
+ y3
)dy =0 (b) (x + y)2
= xy
dy
dx
(c)
1
2x
dy
dx
+
x + y
x2
+ y2
=0 (d) x2
ydx - x3
dy = y3
dy
Homogeneous D.E.