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← JEE Main 2022 · 25 Jul · Shift 2
MathematicsInteger+4 / -1Question 25 of 90

JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 25

Let y=y(x)y=y(x) be the solution of the differential equation
dydx=4y3+2yx23xy2+x3,y(1)=1.\frac{dy}{dx}=\frac{4y^3+2yx^2}{3xy^2+x^3},\quad y(1)=1.
If for some n∈Nn\in\mathbb N, y(2)∈[n−1,n)y(2)\in[n-1,n), then nn is equal to ______.

Answer and worked solution

Official answer: 3

Homogeneous. With y=vxy=vx: (3v2+1)dvv(v2+1)=dxx\dfrac{(3v^2+1)dv}{v(v^2+1)}=\dfrac{dx}x, so v(v2+1)=2xv(v^2+1)=2x.

At x=2x=2. v3+v=4v^3+v=4, v≈1.38v\approx1.38, y≈2.76∈[2,3)y\approx2.76\in[2,3).

Answer: 3