JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

x^{3} d y + (x y - 1) d x = 0, x > 0

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

. Then

y (1)

is equal to

Answer: (A)

Solution

The equation is linear:

\frac{d y}{d x} + \frac{y}{x ^{2}} = \frac{1}{x ^{3}}

, with integrating factor

e^{- 1/ x}

. Then

\frac{d}{d x} (y e^{- 1/ x}) = \frac{e ^{- 1/ x}}{x ^{3}}

, and the substitution

t = - \frac{1}{x}

gives

y e^{- 1/ x} = e^{- 1/ x} (1 + \frac{1}{x}) + C

, so

y = 1 + \frac{1}{x} + C e^{1/ x}

. The condition

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

forces

C = - \frac{1}{e}

. Hence

y (1) = 1 + 1 - \frac{1}{e} \cdot e = 1.

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