If y^{2}+log _{e}left(cos ^{2} xright)=y, x i | vedclass

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(A) Given equation: $y^{2}+\ln(\cos^{2}x) = y$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.At $x=0$,$\cos^{2}(0) = 1$,so $\ln(1) = 0$. The

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$|y^{\prime \prime}(0)|=2$

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(A) Given equation: $y^{2}+\ln(\cos^{2}x) = y$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$.At $x=0$,$\cos^{2}(0) = 1$,so $\ln(1) = 0$. The equation becomes $y^{2} = y$,which implies $y(y-1) = 0$,so $y=0$ or $y=1$.Differentiating with respect to $x$: $2yy^{\prime} + \frac{1}{\cos^{2}x} \cdot 2\cos x \cdot (-\sin x) = y^{\prime}$.Simplifying: $2yy^{\prime} - 2	an x = y^{\prime}$.At $x=0$,for both $y=0$ and $y=1$,we get $2y(0) - 2(0) = y^{\prime}$,which means $y^{\prime}(0) = 0$.Differentiating again: $2y y^{\prime \prime} + 2(y^{\prime})^{2} - 2\sec^{2}x = y^{\prime \prime}$.At $x=0$ and $y^{\prime}(0)=0$: $2y y^{\prime \prime} + 0 - 2(1) = y^{\prime \prime}$.If $y=0$,then $0 - 2 = y^{\prime \prime} \implies y^{\prime \prime}(0) = -2$.If $y=1$,then $2y^{\prime \prime} - 2 = y^{\prime \prime} \implies y^{\prime \prime}(0) = 2$.In both cases,$|y^{\prime \prime}(0)| = 2$.

If $y^{2}+\log _{e}\left(\cos ^{2} x\right)=y, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),$ then

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