What is (\frac{x^{2}+4}{y^{2}+4})\frac{dy}{dx}[(x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y] equal to?

Explanation

From the previous question, (x^2+4)(\frac{dy}{dx})^2 = 16(y^2+4). Differentiating wrt x: 2x(\frac{dy}{dx})^2 + 2(x^2+4)\frac{dy}{dx}\frac{d^2y}{dx^2} = 32y\frac{dy}{dx}. Dividing by 2\frac{dy}{dx}, we get x\frac{dy}{dx} + (x^2+4)\frac{d^2y}{dx^2} = 16y \implies (x^2+4)\frac{d^2y}{dx^2} - 16y = -x\frac{dy}{dx}. Substituting this into the given expression yields (\frac{x^2+4}{y^2+4})\frac{dy}{dx}(-x\frac{dy}{dx}) = -x(\frac{x^2+4}{y^2+4})(\frac{dy}{dx})^2. Since (\frac{dy}{dx})^2 = \frac{16(y^2+4)}{x^2+4}, the expression simplifies to -16x.

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