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Answer

11

Explanation

Given, ` ( a+ bx) e^(y//x) = x rArr y = x log (x/(a+bx)) `
` rArr" " y = x [log (x) - log (a + bx)]` …(i)
On differentiating both sides, we get
`(dy)/(dx) = x (1/x-b/(a+bx))+1 [ log (x) - log (a+bx)]`
` rArr x (dy)/(dx) = x^(2) (a/(x(a+bx)))+y`
` rArr x y_(1) = (a x )/( a + bx) + y ` ...(ii)
Again, differentiating both sides, we get
` xy_(2) + y_(1) = a{((a+bx)* 1 -x*b)/((a+bx)^(2))}+y_(1) `
` rArr" " x^(3) y_(2) = (a^(2)x^(2))/((a+bx)^(2))`
` rArr" "x^(3) y_(2)=((ax)/((a+bx)))^(2)` [from Eq. (ii)]
` rArr" " x^(3)y_(2) = (xy_(1) - y)^(2) `
` rArr" " x^(3) (d^(2)y)/(dx^(2)) = (x(dy)/(dx) - y)^(2)`