JEE Math Practice Question
If the solution $$y=y(x)$$ of the differential equation $$(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0$$ satisfies $$y(-1)=-\frac{\pi}{4}$$, then $$y(0)$$ is equal to :
- A.$$-\frac{\pi}{12}$$
- B.$$\frac{\pi}{2}$$
- C.0
- D.$$\frac{\pi}{4}$$
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