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2
d y f   f dy  f dy f  dy
1  1  1 1  1 2   1 n .
dx 2 x   y dx  y dx  y dx
1 2 n

and in place of y y , , y we put their value from the system
, 
1 2 n
(7.3). As a result we get equation

y  F ( ,x y y , , y ).
,
1 2 1 2 n

Again we differentiate the got equality on x and act like
the above-mentioned ; we get:

y F ( ,x y y , , y ).
,
1 3 1 2 n

So we diyatimo until then, while will not get equation:

,
y ( )n  F ( ,x y y , , y ).
1 n 1 2 n

Then from got on intermediate stages (n  2)- х equations
y  F , y F , , y (n 1)  F and we find equation
1 2 1 3 1 n 1
,
y  f and put in y y , , y the last equation y ( )n  F , as a
1 1 2 3 n n
result will obtain equation:

,
y ( )n  F * ( ,x y y , , y (n 1) ).
1 1 1 1

We decide him and will get a function
)
y   * ( ,x C ,C , ,C .
1 1 1 2 n
,
We search other unknown functions y y , , y by
2 3 n
auxiliary equations, which turned out on intermediate stages.

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