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Differential Equations question

2022 · 25 Jul · Shift 2 · Q41
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  5. /2022 · 25 Jul · Shift 2 · Q41

Differential Equations question

2022 · 25 Jul · Shift 2 · Q41

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx=4y3+2yx23xy2+x3,y(1)=1\frac{d y}{d x}=\frac{4 y^{3}+2 y x^{2}}{3 x y^{2}+x^{3}}, y(1)=1dxdy​=3xy2+x34y3+2yx2​,y(1)=1. If for some n∈N,y(2)∈[n−1,n)n \in \mathbb{N}, y(2) \in[n-1, n)n∈N,y(2)∈[n−1,n), then nnn is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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