JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

\frac{d y}{d x} = \frac{4 y ^{3} + 2 y x ^{2}}{3 x y ^{2} + x ^{3}}, y (1) = 1

. If for some

n \in N, y (2) \in [n - 1, n)

, then

n

is equal to

Answer: 3

Solution

The equation is homogeneous, so substituting

y = v x

reduces it to

x \frac{d v}{d x} = \frac{v ( v ^{2} + 1 )}{3 v ^{2} + 1}

, that is

\frac{3 v ^{2} + 1}{v ( v ^{2} + 1 )} d v = \frac{d x}{x}

. Splitting the left side as

\frac{1}{v} + \frac{2 v}{v ^{2} + 1}

and integrating gives

ln (v (v^{2} + 1)) = ln x + const

, hence

v (v^{2} + 1) = K x

. Returning to

y

this is

y (x^{2} + y^{2}) = K x^{4}

, and the condition

y (1) = 1

fixes

K = 2

, so

y (x^{2} + y^{2}) = 2 x^{4}

. At

x = 2

this reads

y^{3} + 4 y = 32

, whose real root is

y \approx 2.76

. Since this lies in

[2, 3)

, the required integer is

n = 3

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer