JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Differential Equations

Let

y = f (x)

be the solution of the differential equation

y (x + 1) d x - x^{2} d y = 0, y (1) = e

. Then

lim_{x \to 0^{+}} f (x)

is equal to

Answer: (A)

Solution

Separating variables in

y (x + 1) d x = x^{2} d y

gives

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

, so

ln y = ln x - \frac{1}{x} + C

. The condition

y (1) = e

gives

1 = 0 - 1 + C

, so

C = 2

and

y = x e^{2 - 1/ x}

. As

x \to 0^{+}

- \frac{1}{x} \to - \infty

makes

e^{2 - 1/ x} \to 0

far faster than

x \to 0

, so the product tends to

0

. The limit is

0

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