Question

For the following differential equation find the particular solution satisfying the given condition:

`y(1 + log x) dx/dy - x log x = 0, y = e^2,` when x = e

Answer 1

`y(1 + log x) dx/dy - x log x` = 0

∴ `(1 + log x)/(x log x)dx - dy/y` = 0

Integrating both sides, we get

∴ `int (1 + log x)/(x log x)dx - dy/y` = c₁      .....(1)

Put x log x = t

Then `[x * d/dx (log x) + (log x) * d/dx (x)]dx` = dt

∴ `[x/x + (log x)(1)]dx` = dt

∴ `(1 + log x)dx` = dt

∴ `int (1 + log x)/(x log x)dx = intdt/t = log |t| = log |x log x|`

∴ From (1), the general solution is

log |x log x| – log |y| = log c, where c₁ = log c

∴ log `|(x log x)/y|` = log c

∴ `(x log x)/y` = c

∴ x log x = cy

This is the general solution.

Now, y = `"e"^2`, when x = e

∴ e log e = c.e²

∴ 1 = c.e     ...[∵ log e = 1]

∴ c = `1/e`

∴ The particular solution is x log x = `(1/e)y`

∴ y = ex log x.