Page 124 - DMTH202_BASIC_MATHEMATICS_II

Unit 9: Solution of Differential Equation

Integrating, we get Notes
1 2y 1 3x x 3
e  e   c
2 3 3

or 3e 2y   2 e 3x  x 3   c , where c 1  6c
1
which is the required solution.

Example:

2 dy
Solve    1x y   1
dx
Solution:
Putting    1x y   u

dy du
we get 1 
dx dx

dy du
or   1
dx dx
Thus the given equation reduces to

u 2  du  1    1

 dx 

du 1  u 2
or  2
dx u
u 2
or 2 du  dx
1  u

 1 
or  1  2  du  dx
 1  u 
Integrating, we get

x
u tan 1 u   c
or    1x y   tan 1   1x y   x c

,
or y  tan 1    1x y   c where c = c – 1
1
1
which is the required solution.
Remark:
Different equations of the form

dy

   ax by c  …..(1)

dx

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