If y2 + loge (cos2x) = y, [x ( - 2, 2 )], then :

Question

If y2 + loge (cos2x) = y,
$$x \in \left( { - {\pi  \over 2},{\pi  \over 2}} \right)$$, then :

Accepted Answer

Given y2 + loge (cos2x) = y .....(1)

Put x  = 0, we get

y2 + loge (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

Solutions · 1