Page 150 - DMTH202_BASIC_MATHEMATICS_II

Unit 11: Linear Differential Equations of First Order

 The solution of (1) is Notes
4x 2 2

2
y .  1 x    .  1 x dx c
1 x 2
4 3
2
c
or y .  1 x   x  .
3
which is the required solution.
!

Caution The system is said to be consistent if it contains a solution or else the system is said
to be inconsistent.

Example:

1

Solve  1 y 2  dx   tan y x dy .
Solution:
-1
Since the equation involves a term tan dy. So it is not linear in y. However, the equation is linear
in x, so that

 1 y 2  dx xdy  tan 1 y dy .....(1)
1
dx 1 tan y
or  2 x  2
dy 1  y 1  y
1
1 tan y
Here P  2 , Q  2 .
1  y 1  y

1
 dy
 I . .  e Pdy  e 1 y 2  e (tan 1 ) y .
F
 The solution of (1) is
1
xe (tan 1 ) y   tan y . e  tan 1  y dy  c .....(2)
1  y 2

Now solving

tan 1  y 1
I   e tan ydy
1  y 2
Putting t = tan-1y

1
dt = 2 dy
1  y
t
or I  te dt
= te – e t
t

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