Page 60 - 4549

Example 3.6 To find the general integral of differential

equation 1 y 2  y y .  
dp

 
 We will lay down y  p (y ), y  . p Equation
dy
dp
2
1 p  yp ;this equation of the first order in relation to р(у)
dy
with the separated variables will collect a kind . We separate
variables and integrate:
pdp dy 1 d (p 2  )1 dy 2
 ;    ; ln( p  )1  ln2 y  ln2 c ;
p 2 1 y 2 p 2 1 y 1

p 2  1  c 2 y 2 ; p   c 2 y 2  . 1
1 1

From here, going back to an old unknown function y , we
will get:
dy 1 d (c ) y
 y   c 1 2 y 2  ,1  dx ,  1   dx .

c 2 y 2 1 c 1 (c ) y 2 1
1 1
It is possible to write down the general integral of the given
equation in a kind:

dy 1 d (c ) y
 y   c 2 y 2  ,1  dx ,  1   dx .

1 2 2 2
c y 1 c 1 (c ) y 1
1 1


Example 3.7 To solve Cauchy task
3
3
y y    y 1  , 0 y ) 1 (  , 1  y ) 1 (  3 .
2

58