Page 142 - DMTH202_BASIC_MATHEMATICS_II

Unit 10: Homogeneous Equations

Notes
Example:

dy 6x  2y  7
Solve  ….(1)
dx 2x  3y  6
Solution:
Putting x = X – h , y – Y + k,

k

dy 6  x h   2(y  ) 7
  ,
h

k
dx 2(x  ) 3(y  ) 6

dY 6X  2Y  6h  2k 7
or  , ….(2)
dX 2X  3Y  2h  3k  6
Choosing h and k such that (2) becomes homogeneous,
i.e., 6h – 2k – 7 = 0 and 2h + 3k – 6 = 0
3
which gives h  , k = 1.
2
 The equation (2) becomes

dY 6X  2Y

dX 2X  3Y
dY dv
v
Putting Y = vX, so that   X .
dX dX
dv 6  2v
 v  
dx 2  3v
dx 1 6v  4
or   2 dv
x 2 3v  4v  6
Integrating, we have

1
log X   log  2v 2  4v  6   logc
2
1
or log x  log  3v 2  4v   6 2  logc

or X 3v 2  4v  c
6

1
or x  3v 2  4v   6 2  c

1
   2    2
 3   3  y  1   y  1     c
or  x      x 3     4  x 3    6 
2

  2   2  
 

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