JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Differential Equations

If a curve

y = f (x)

passes through the point

(1, 2)

and satisfies

x \frac{d y}{d x} + y = b x^{4}

, then for what value of

b, \int_{1}^{2} f (x) d x = \frac{62}{5}

Answer: (C)

Solution

The equation

x \frac{d y}{d x} + y = b x^{4}

is exactly

\frac{d}{d x} (x y) = b x^{4}

, so

x y = \frac{b x ^{5}}{5} + C

. The point

(1, 2)

gives

C = 2 - \frac{b}{5}

, hence

f (x) = \frac{b x ^{4}}{5} + \frac{1}{x} (2 - \frac{b}{5})

and

\int_{1}^{2} f (x) d x = [\frac{b x ^{5}}{25}]_{1}^{2} + (2 - \frac{b}{5}) ln 2 = \frac{31 b}{25} + (2 - \frac{b}{5}) ln 2.

As a function of

b

this is linear with slope

\frac{31}{25} - \frac{ln 2}{5} > 0

, so at most one

b

can make it

\frac{62}{5}

; and

b = 10

does, since then the logarithmic term vanishes and

\frac{310}{25} = \frac{62}{5}

. The required value is

b = 10

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option