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

` y(1+ log x) (dx)/(dy) - x log x =0`
` (1 +log x)/(x log x) dx - (dy)/(y) =0`
`int (1+ log x)/(x log x) dx - int (dy)/y =0`
put x log x=t
Then ` [ x. d/(dx) l(log x) +(log x) . d/(dx) (x) ] dx=dt `
` [ x/x + (log x) (1)] dx = dt " " therefore (1+ log x ) dx = dt `
`int (1 + log x)/(x log x) dx = int (dt)/t = log |t| = log |t| log |x log x|`
from (1), the general solution is
` log |x log x| - log |y| = log c, " where" c_(1) = log c`
` log |( x log x)/y| = log c " " therefore (x log x) / y = c`
x log x = cy
Now ` , y =e^(2) `, when x =e
` e log e = c.e^(2) " " 1 =c.e `
` c =1 /e`
the particular solution is x log x = ` (1/e)y `
y = ex log x