If a neq b, x neq n pi, n in Z and y^2 = a^2 | vedclass

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(B) Given $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$. Differentiating with respect to $x$: $2y \frac{dy}{dx} = -2a^2 \cos x \sin x + 2b^2 \sin x \cos x = (b^

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$\frac{1}{y} \left( \frac{ab}{y} \right)^2$

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(B) Given $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$. Differentiating with respect to $x$: $2y \frac{dy}{dx} = -2a^2 \cos x \sin x + 2b^2 \sin x \cos x = (b^2 - a^2) \sin 2x$. Thus,$y \frac{dy}{dx} = \frac{b^2 - a^2}{2} \sin 2x$. Differentiating again with respect to $x$: $y \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = (b^2 - a^2) \cos 2x$. From the first derivative,$\frac{dy}{dx} = \frac{(b^2 - a^2) \sin 2x}{2y}$. Substituting this into the second derivative equation: $y \frac{d^2 y}{dx^2} + \frac{(b^2 - a^2)^2 \sin^2 2x}{4y^2} = (b^2 - a^2) \cos 2x$. Multiplying by $y^2$: $y^3 \frac{d^2 y}{dx^2} + \frac{(b^2 - a^2)^2 \sin^2 2x}{4} = y^2 (b^2 - a^2) \cos 2x$. Using $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$,we find that $y^3 (\frac{d^2 y}{dx^2} + y) = a^2 b^2$. Therefore,$\frac{d^2 y}{dx^2} + y = \frac{a^2 b^2}{y^3} = \frac{1}{y} \left( \frac{ab}{y} \right)^2$.

If $a \neq b, x \neq n \pi, n \in Z$ and $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$,then $\frac{d^2 y}{dx^2} + y =$

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