JEE Math Practice Question
Let $y=y(x)$ be the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$. Then, $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ equals :
- A.$\frac{4}{4+\sqrt{\mathrm{e}}}$
- B.$\frac{3}{3-\sqrt{\mathrm{e}}}$
- C.$\frac{2}{1+\sqrt{\mathrm{e}}}$
- D.$\frac{1}{2-\sqrt{\mathrm{e}}}$
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