JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Differential Equations

If the solution curve of the differential equation

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

passes through the points

(e, \frac{4}{3})

and

(e^{4}, α)

, then

α

is equal to ______.

Answer: 3

Solution

Rewrite the differential equation as:

\frac{d y}{d x} + \frac{y}{2 x ln x} = \frac{1}{x}

This is a first-order linear ODE. The integrating factor is:

I.F. = e^{\int \frac{1}{2 x l n x} d x} = e^{\frac{1}{2} l n (l n x)} = ln x

Multiplying through and integrating:

y ln x = \int \frac{ln x}{x} d x

Substitute

ln x = t^{2}

, so

\frac{d x}{x} = 2 t d t

y ln x = \int 2 t^{2} d t = \frac{2 t ^{3}}{3} + C = \frac{2 ( ln x ) ^{3/2}}{3} + C

Using the point

(e, 4/3)

\frac{4}{3} \cdot 1 = \frac{2}{3} + C

, so

C = \frac{2}{3}

.

The solution is

y ln x = \frac{2 ( l n x ) ^{3/2}}{3} + \frac{2}{3}

.

At

(e^{4}, α)

α \cdot 2 = \frac{2 \cdot 8}{3} + \frac{2}{3} = 6

, so

α = 3

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Integer answer