To solve it we make the substitution
y = vx (3)
Differentiating equation (3) with respect to, x we get
dy
dx  =
dv
v
+x dx (4)
Substituting the value of y  and dy
dx in equation (1) we get
dv
v
+x dx
 = 1
12
+vv
−
or
dv
x dx  = 1
12
v
v
+v
−
−
or
dv
x dx  =
2
1
v1
vv
 
or
2
1
1
v
dv
v
v  
 =
dx
x
 
Integrating both sides of equation (5), we get
2
1
1
v
dv
v
 v
 
 
 =
xdx
 
or
2
1
2
1
3
2
1
v
dv
v
 v
 
 
 
 
 = – log |x| + C1
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DIFFERENTIAL EQUATIONS
401
or
1
2
2
1
2
1
3
1
log
C
2
2
1
1
 
  
 
 
 
 
 
 
 
v
dv
dv
x
v
v
v
v
or
2
1
2
2
1
3
1
log
1
log
C
2
2
1
3
2
2
 
 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
v
v
dv
x
v
or
2
1
1
1
3
2
2
1
log
1 tan
log
C
2
2
3
3
 
 
 
 
 
 
  
 
 
 
 
 
v
v
v
x
or
2
2
1
1
1
1
2
1
log
1
log
3 tan
C
2
2
3
 
 
 
 
 
 
 
 
 
 
 
 
v
v
v
x
(Why