Answer
$y = k(16-x^2)^2 + 2$
Work Step by Step
Given that:
$\frac{dy}{dx} = \frac{x^2y - 32}{16 - x^2} + 2$
Add terms on the right side into a single fraction and simplify:
$$\frac{dy}{dx} = \frac{x^2y -32}{16 - x^2} + \frac{2(16 - x^2)}{16 - x^2}$$
$$\frac{dy}{dx} = \frac{x^2y -32 + 32 - 2x^2}{16 - x^2}$$
$$\frac{dy}{dx} = \frac{x^2y - 2x^2}{16 - x^2}$$
$$\frac{dy}{dx} = \frac{x^2(y - 2)}{16 - x^2}$$
Separate and integrate:
$$\frac{dy}{y-2} = \frac{x^2}{16 - x^2}dx$$
$$\int{\frac{dy}{y-2}} = \int{\frac{x^2}{16 - x^2}}dx$$
$$\ln{|y-2|} = \int{\frac{x^2}{(4-x)(4+x)}}dx$$
$$\ln{|y-2|} = \int{\frac{2}{4-x}+\frac{2}{4+x}}dx$$
$$\ln{|y-2|} = 2\ln{|4-x|}+2\ln{|4+x|} +C$$
$$\ln{|y-2|} = \ln{(16-x^2)^2} +C$$
Solve for $y$:
$$y-2 = e^{\ln{(16-x^2)^2} +C}$$
$$y-2 = (16-x^2)^2 \cdot e^C$$
Let $e^C = k$:
$$y-2 = k(16-x^2)^2$$
$$y = k(16-x^2)^2 + 2$$
