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d cos x 1
  cos2 x   c 1 cos x  cos2 x  c 1 cosx  2  c 1 cos . x
2
cos x cos x

As, we come to differential equation z  y  of the 1-st
type, which easily gets untied by double integration:

,
  y  2  с cos , x  y  ( 2  с cos x )dx  2x  с sin x  c
1  1 1 2
2
y   2 ( x  с 1 sin x  c 2 )dx  x  c 1 cos x  c 2 x  c 3 .

Thus, the sought common decision after of the given
equation of 3-rd order was got:
2
y  x  c cos x  c x  c . 
1 2 3

Example 3.4 To find the partial decision of equation
y
y x   y ln , ) 1 ( y  e , y ) 1 (  e 2 .
x

 We have equation of 2-nd type (n=2, k=1), as it does not
contain y. We will bring an order down of this equation on 1,
 
laying down z  . y Then y  , z and the given equation
grows into homogeneous differential equation of the first order
of relatively unknown function of z:
z z z


zx  z ln or z  ln .
x x x
z
We decide it by substitution u  , z  ux , then
x

z  u x  , u and the given equation collects a kind
u x  u  u ln . u
Separating variables and integrating, we find consistently:
du dx d (lnu  )1 dx
 ,    , ln lnu 1  ln x  lnc ,
u (lnu  )1 x lnu 1 x 1
ln u 1  c , x ln  cu x  , 1 u  e c 1 x 1  , z  xe c 1 x 1 .
1 1
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