Page 138 - DMTH202_BASIC_MATHEMATICS_II

Unit 10: Homogeneous Equations

Integrating, we get Notes


log v  1 v 2  log x log c
or v  1 v 2  cx


y y 2  y 
or  1  2  cx  v  
x x  x 

or y  x 2  y 2  cx 2

which is the required solution.
   2  y   2 y
y
 
x
y
Example: Solve  x tan    sec   dx  sec   dy  0
x
x
x
      
Solution:
2 y  
y
 
y sec    tan  
x
dy    
x
x

Here dx 2 y …..(1)
 
x sec  
x
 
y
Putting v i.e., y = vx
x
dy dv
v
So that   x
dx dx
Equation (1) reduces to
dy tan v
v  x  
v
2
dx sec v
2
sec v dx
or dv   0
tanv dx
Integrating, we get
log (tan v) + log x = log c
or log (x tan v) = log c
y
    y
c
or x tan    ;   v  
x
    x
which is required solution.
 x y  x y  x 
Example: Solve  1  e  dx  e  1   dy  0 .
   y 

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