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Differential Equations question

2024 · 1 Feb · Shift 1 · Q44

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JEE Main
Mathematics
Differential Equations
2024 · 1 Feb · Shift 1 · Q44

Differential Equations question

2024 · 1 Feb · Shift 1 · Q44

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

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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