Page 156 - DMTH202_BASIC_MATHEMATICS_II

Unit 11: Linear Differential Equations of First Order

Equation (1) becomes Notes
dz 3
  yz y

dy
dz 3
or  yz   y which is linear in z.
dy

 y dy y 2
I.F.  e  e 2 .
 The solution is

y 2 3 y 2
Z.e 2   y .e 2 dy  c

2
   y .e y 2 2 .y dy  c

2
t  y 

   2t.e dt c, t  

 2 
t

t
   2 .e  e t   c
   
y 2 y 2  y 2 
Ze 2   2e 2    c
1 
 2 
1  y 2   y 2 2
1 
   2    ce
x  2 
which is the required solution.

Example:

dy 2 3 2
Solve  xsin y  x cos y .
dx
Solution:
2
Dividing by cos y, we have
y
dy 2 sin cosy
x
2
sec y .   x 3
2
dx cos y
dy 3
2
y
or sec y  2tan .x  x . …..(1)
dx
Putting tan y = z
dy dz
2
 sec y  .
dx dx

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