The particular solution of the differential equation ` y (1+logx)(dx)/(dy)-xlog x =0 ," when" x = e, y = e^(2)` is

Options

A. `y = ex log x`
B. ` ey = x log x`
C. ` xy = e log x`
D. ` y log x = ex`

Answer

a

Explanation

Given , differential equation is
` y(1+logx)(dx)/(dy) - x log x =0`
` rArr ((1+logx)dx)/(x log x)=(dy)/y`
` rArr (1/(xlogx)+1/x)dx = 1/y dy`
On integrating both sides , we get
` int (1/(xlogx)+1/x)dx= int 1/y dy `
Put log ` x = t rArr 1/x dx = dt `
` :. int 1/t dt +int 1/x dx = int 1/y dy `
` rArr log t + log x = log y +log c`
` rArr log tx = log yc`
` rArr t x = yc rArr x log x = yc`
When x = e and `y = e^(2)`
` :. e log e = e^(2) c`
` rArr 3 xx 1 = e^(2) c rArr c rArr 1/e`
` :. x log x = y/e rArr y = ex " log " x `