Given y² + log_e (cos²x) = y .....(1)

Put x = 0, we get

y² + log_e (1) = y

$$ \Rightarrow $$ y² = y

$$ \Rightarrow $$ y = 0, 1

Differentiating (1) we get

2yy' + $${1 \over {\cos x}}\left( { - \sin x} \right)$$ = y'

$$ \Rightarrow $$ 2yy' - 2tanx = y' ....(2)

From (2) when x = 0, y = 0 then y'(0) = 0

From (2) when x = 0, y = 1 then

2y' = y'

$$ \Rightarrow $$ y'(0) = 0

Again differentiating (2) we get

2(y')² + 2yy'' – 2sec²x = y''

from (2) when x = 0, y = 0, y’(0) = 0 then

y”(0) = -2

Also from (2) when x = 0, y = 1, y’(0) = 0 then

y”(0) = 2

$$ \therefore $$ |y''(0)| = 2