Find the particular solutions of the following differential equation :
(1) ` y(1+ log x) dx/dy - x log x = 0 , when , x =e , y= e^(2)`

Explanation

Given differential equation is ,
` y (1 + log x) * (dy)/(dx) - x * log x = 0`
` :' " " (1 + log x) * dx * y/(dy) = x * log x `
` :. " " (1 + log x)/(x * log x) * dx = (dy)/y`
Integrating both sides,
` int (1+log x)/(x * log x) * dx = int(dy)/y `
Put ` x * log x = t`
` :. (x * 1/x + log x * 1) * dx = dt `
` rArr " " (1 + log x ) * dx = dt`
` rArr int (dt)/t = int (dy)/y`
` rArr log t = log y + log C `[log C - constant of integration]
` rArr" " log (x * log x) = log (y * C) `
` rArr x * log x = y *C` ...(1)
which is the solution of differential equation.
For particular , put `y = e^(2) ` and x = e .
` :. ` Equation (1) becomes
` e * log e = e^(2) * C`
`rArr" " e * 1 = e^(2) * C`
` rArr" " C = e/e^(2) = 1/e`
` rArr" " C = 1/e`
` :. ` Equation (1) becomes
` x * log x = y * 1/e`
` :. " " y = e * x * log x `
This is the required particular solution.