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1) to work out a characteristic equation (5.3) ;
2) to find the roots of the got characteristic equation;
3) after the roots of characteristic equation to write down n
linearly independent decisions equation (5.1):
x 
а) partial decision answers every simple root – e ;
б) if  – real root of multiple S, S answers him linearly
x 
independent decisions – eC x  , хe ,..., x s 1 e  x ;
1
в) every steam complex conjugating roots    і  ,  і 
answers two linear independent partial decisions –
x


x
e cos x  ; e sin x  ;
г) to every steam complex conjugating roots    і  ,  і 
of multiple of к answers 2к linear independent partial decisions
– (5.16).
These partial decisions will be exactly so much, what
degree of characteristic equation( that is so much, what order of
differential equation). These decisions are linear independent.
4) finding n linear independent decisions y , y , y ,.....y ,
1 2 3 n
general decision of equation (5.1):

y  C y  C y  C y .....  C y . (5.18)
1 1 2 2 3 3 n n

Example 5.4 To find the common decision of equation
    y    y   y y  0 .
 We make characteristic equation  3   2    1 0
 2 (  )1  (  )1  0
(  )1 ( 2  )1  0

   ;1   i 
1 3 , 2

partial decisions will be written down as:

y  e  x y ;  cos x; y  sin x .
1 2 3

The common decision is such therefore :

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