Page 143 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes which on simplification gives
3y + 4xy – 6x – 12y + 14x = c.
2
2
Example:
dy 6x  2y  7
Solve  …(1)
y
dx 3x   4
Solution:

a b
Here  .
A B
dy dv
So putting 3x – y = v, so that 3  ,
dx dx
 (1) becomes

dv 2v 7 v  19
  
3
dx v  4 v  4
v  4  15 

 dx  dv  1   dv
v  19  v  19 
Integrating, we get

x + c = v – 15 log(v + 19)
On restoring the value of v, we get
2x – y – 15 log (v + 19) = c,
which is the required solution.

Example:

dy 2x  3y  4
Solve  , given that y = 1, when x = 1.
y
dx 4x   3
Solution:

Putting x = X + h and y = Y + k, the given equation becomes
dY  2 X h   3  Y k  4

 ,
dX  4 X h   Y k  3


2X  3Y  2h  3k  4
 ….(1)
k
4X Y  4h   3
Choosing h and k such that (1) becomes homogeneous,
i.e., 2h + 3k – 4 = 0 and 4h + k – 3 = 0.

1
 h  ,k  1.
2

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