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If y^2=a^2cos^2x+b^2sin^2x then prove th...

If `y^2=a^2cos^2x+b^2sin^2x` then prove that `(d^2y)/(dx^2)+y=(a^2b^2)/(y^3)`

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`y^{2}=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x`
Differentiating both sides w.r.t. `x`, we get
`2 y \frac{{dy}}{{dx}}=a^{2} \frac{{d}}{{dx}}(\cos x)^{2}+b^{2} \frac{{d}}{{dx}}(\sin x)^{2} =a^{2} \times 2 \cos x \cdot \frac{{d}}{{dx}}(\cos x)+b^{2} \times 2 \sin x \cdot \frac{{d}}{t} d x(\sin x) =a^{2} \times 2 \cos x(-\sin x)+b^{2} \times 2 \sin \times \cos x =(b^{2}-a^{2}) \sin 2 {x} \therefore y \frac{{dy}}{{dx}}=(\frac{b^{2}-a^{2}}{2}) \sin 2 x`
Differentiating again w.r.t. `x`, we get ...
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