Page 144 - DMTH202_BASIC_MATHEMATICS_II

Unit 10: Homogeneous Equations

Notes
dY 2X  3Y
Then (1) becomes  .
df 4X Y
dY dv
v
Putting Y = vX, so that   X .
dX dX
dv 2  3v
 v  X 
dX 4  v

v
dv 2  3v 2   v 2
v
or X   
dX 4  v 4  v
v  4 dX
or dv  
  1v   2v  X

 2 5  1
or     dv   dx
 3   2v  3   1v   x
Integrating, we have

2 5
 log   2v   log   1v   log X  logC
3 3
Y
Replacing v by , we have
X

 Y  2X   Y  X 
3logX  3logC  2log    5log  
 X   X 
On restoring the value of x & y, we get
 1 
5log y  x    2log   2y x  2   3log c .

 2 

 1  5
 y x
  
or  2  2  c 3 ;
 2y x   2
1
3
Putting y = 1 at x = 1, we have c   .
25
 1  5
 y x
  
Thus  2    1
(y  2x  2) 2 2 5
1
5
2
or (y+2x–2) = –2 (y–x  ) 5
2
= (2x–2y+1) 5
2
or (y+2x–2) = (2x–2y+1) 5
which is the required particular solution.

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