Question

Find the particular solution of the following differential equation:

y(1 + log x) = (log x^x) `"dy"/"dx"`, when y(e) = e²

Answer 1

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

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

`y(1 + log x)dx/dy - xlogx = 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" = "c"_1`      .....(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"`

∴ `int (1 + log "x")/("x" log "x")"dx" = int"dt"/"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

∴ c = `1/"e"`

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

∴ y = ex log x.