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

y = y (x)

is the solution of the differential equation

2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0

such that

y (e) = \frac{e}{3}

, then

y (1)

is equal to

Answer: (B)

Solution

Rewrite the differential equation as

\frac{d y}{d x} - \frac{y}{x} = - \frac{3}{2} (\frac{y}{x})^{2}

. Use the substitution

y = v x

, so

\frac{d y}{d x} = v + x \frac{d v}{d x}

. Substituting gives

v + x \frac{d v}{d x} - v = - \frac{3}{2} v^{2}

, which simplifies to

x \frac{d v}{d x} = - \frac{3}{2} v^{2}

. Separating variables:

\frac{d v}{v ^{2}} = - \frac{3 d x}{2 x}

. Integrating:

- \frac{1}{v} = - \frac{3}{2} ln ∣ x ∣ + C

, or

- \frac{x}{y} = - \frac{3}{2} ln ∣ x ∣ + C

. Using the initial condition

y (e) = \frac{e}{3}

- 3 = - \frac{3}{2} + C

, so

C = - \frac{3}{2}

. The solution is

- \frac{x}{y} = - \frac{3}{2} ln ∣ x ∣ - \frac{3}{2}

. At

x = 1

- \frac{1}{y} = - \frac{3}{2}

, so

y (1) = \frac{2}{3}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option