Post on 27-Apr-2015
description
HOMOGENEOUS DIFFERENTIAL
EQUATIONS
Homogeneous FunctionA function f(x,y) is called Homogeneous of degree n if
Where t is a nonzero real number. Thus
are
Homogeneous function of degree 1, 8 and 0 respectively
),(),( yxftyxf n
y
xand
yx
yxxy sin..., 22
1010
Homogeneous Equation
A first order DE of the form
Is said to be Homogeneous if the function f does not depend on x and y separately, but only on ratio . Thus first order homogeneous
equation are of the form ---------(1)
A homogeneous equation
Is transformed into a separable equation (in the variables y and x) by the substitution y = vx
),( yxfdx
dy
x
y
x
yg
dx
dy
x
yg
dx
dy
Put y = vx and in eq (1)
This can be separated and be solved
)(vgdx
dvxv
dx
dvxv
dx
dy
0)( dx
dvxvgv
0)]([ xdvdxvgv
x
dx
vgv
dv
)(
02 22 dyxxyy
2
2 2
x
xyy
dx
dy
vxy Put and
Solve
Soln:
dx
dvxv
dx
dy
-----------(1)
So eq (1) becomes
vvx
xvxxv
dx
dvxv 2
2 22
22
vvdx
dvx 32
x
dx
vv
dv
3
cxvv loglog3log3
1log3
1
x
dxdv
vv 3
11
3
1
x
c
v
v 1log33
log
3
1log3
log
x
c
v
v
33
13
3 x
c
x
c
v
v
3
3 x
c
xyxy
)3(3 ycyx
0)4(52 dyyxdxyx
)4(
)52(
yx
yx
dx
dy
vxy Put and
Solve
Soln:
dx
dvxv
dx
dy
when y(1)=4
So eq(1) becomes
)4(
)52(
)4(
)52(
v
v
vxx
vxx
dx
dvxv
vv
v
dx
dvx
)4(
)52(
x
dx
vv
dvv
2)1(
)4(
-----------(1)
cxvv loglog2log2)1log(
x
dx
v
dv
v
dv
22
1
cxv
vlog
)2(
)1(log 2
2]2[]1[ x
ycx
x
y 2]2[][ xycxy
cxv
v
2)2(
)1(
2]24[]14[ c 2]24[]14[ c
12
1 c 2]2[][12 xyxy
dxyxxdxydy 22
Solve
Solve
1)1(....0)( 22 ywhendyxdxyxy
EQUATIONS REDUCIBLE TO HOMOGENEOUS
FORM
Equation Reducible to Homogeneous FormThe DE
Is not homogeneous. It can be reduced to homogeneous form as explained below
Case-I If then make the
transformation x = X + h, y = Y + k
0)()( 222111 dycybxadxcybxa
2
1
2
1
b
b
a
a
0)(
)(
22222
11111
dYckbhaYbXa
dXckbhaYbXa
Let h and k be the solution of the system of equations
Then for calculated values of h and k eq (1) will be reduced to homogeneous form
In the variables X and Y
0
0
222
111
ckbha
ckbha
0)()( 2211 dYYbXadXYbXa
Case-II If
then put
And the given equation will reduce to a separable equation in the variables x and z
ybxaz 11 2
1
2
1
b
b
a
a
Solve
Soln: Let x = X+h and y = Y+k, then
Now 5h + 5 = 0 h = -1 k = 1
32
12
yx
yx
dx
dy
3)(2
1)(2
kYhX
kYhX
dx
dy
322
122
khYX
khYX
dx
dy
032
0122
kh
kh
Put Y = vX
3212
1122
YX
YX
dx
dy
YX
YX
dx
dy
2
2
XYXY
dx
dy
21
2
v
v
dX
dvXv
21
2
vv
v
dX
dvX
21
2
cXvv lnln2)1ln(tan 21
v
v
v
vvv
dX
dvX
21
)1(2
21
22 22
X
dXdv
v
v2
1
)21(2
X
dX
v
vdv
v
dv2
1
2
1 22
cXvv lnln)1ln(tan 221
221 )1(lntan Xvcv
22
21 1lntan X
X
Yc
X
Y
)(lntan 221 YXcX
Y
])1(
)1[(ln)1(
)1(tan
2
21
y
xcx
y
Solve
Soln: Let z = 3x – 4y then
343
243
yx
yx
dx
dy
dx
dy
dx
dz43 dx
dz
dx
dy)4
1(
4
3
3
2)4
1(
4
3
z
z
dx
dz
3
2
4
3)4
1(
z
z
dx
dz
)3(4
)1()4
1(
z
z
dx
dz
)3(
)1(
z
z
dx
dz
dx
z
dzz
)1(
)3(
dxz
dzdz
)1(4
)1ln(41 zcxz
)143ln(443 1 yxcxyx
)143ln(41 yxc
yx
)143ln( yxcyx
Put z = 3x – 4y
Solve
5
1
xy
xy
dx
dy Solve
12
52
yx
yx
dx
dy