Question

Find the particular solution of the differential equation:

`y(1+logx) dx/dy - xlogx = 0`

when y = e² and x = e

Answer 1

Given equation is

`y(1 + logx) dx/dy -xlogx = 0`

`:. y(1+logx) dx/dy = xlogx`

`:. y(1+logx)dx = xlogx dy`

Separating the variables

`1/ydy = (1+logx)/(xlogx) dx`

Integrating, we have

`int1/y dy = int (1+logx)/(xlogx) dx`

`:.log|y| = log|xlogx|+logc`

`:. log|y| = log|cxlogx|`

∴ y = cx log x is the general solution

Given x = e,    y = e²

∴ e² = c.e.log e

∴ `e^2 = c.e`

∴ c = e

∴y = ex.logx