JEE Math Practice Question
Let $$y=y(x)$$ be the solution of the differential equation $$(x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0$$. If $$y(0)=0$$, then $$y(2)$$ is equal to
- A.$$2 \pi$$
- B.$$\frac{\pi}{8}$$
- C.$$\frac{\pi}{16}$$
- D.$$\frac{\pi}{32}$$
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