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

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

$\frac{ab}{y^{3}}$

${(\frac{ab}{y})}^{2}$

$\frac{{(ab)}^{2}}{y}$

answer is B.

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Detailed Solution

$\begin{array}{l} y^{2} = a^{2} \cos^{2} x + b^{2} \sin^{2} x \\ 2 y^{2} = a^{2} (2 \cos^{2} x) + b^{2} (2 \sin^{2} x) \\ 2 y^{2} = a^{2} (1 + \cos^{2} x) + b^{2} (1 - \cos^{2} x) \\ 2 y^{2} = (a^{2} + b^{2}) + (a^{2} - b^{2}) \cos^{2} x \end{array}$
$\begin{aligned} 4 y \frac{d y}{d x} = 2 (a^{2} - b^{2}) \sin 2 x \\ 2 y \frac{d y}{d x} = (a^{2} - b^{2}) \sin 2 x \end{aligned} \begin{array}{l} y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = (a^{2} - b^{2}) \cos 2 x \\ y^{3} \frac{d^{2} y}{d x^{2}} + y^{2} {(\frac{d y}{d x})}^{2} = y^{2} (a^{2} - b^{2}) \cos 2 x \\ y^{3} \frac{d^{2} y}{d x^{2}} = y^{2} (a^{2} - b^{2}) \cos 2 x - y^{2} {(\frac{d y}{d x})}^{2} \end{array}$
$y^{4} + y^{3} \frac{d^{2} y}{d x^{2}} = y^{4} + y^{2} (a^{2} - b^{2}) \cos 2 x - y^{2} (\frac{d y}{d x}) y^{2} (y^{2} + (a^{2} - b^{2})) \cos 2 x - (b^{2} - a^{2}) \sin^{2} x \cos^{2} x$ $a^{2} \cos^{2} x + b^{2} \sin^{2} x) (b^{2} \cos^{2} x + a^{2} \sin^{2} x) - (b^{2} - a^{2}) \sin^{2} x \cos^{2} x \begin{array}{l} a^{2} b^{2} (\cos^{4} x + \sin^{4} x + 2 \sin^{2} x \cos^{2} x) \\ a^{2} b^{2} {(\sin^{2} x + \cos^{2} x)}^{2} = a^{2} b^{2} \\ y + \frac{d^{2} y}{d x^{2}} = \frac{a^{2} b^{2}}{y^{3}} \end{array}$