The general solution of `x((dy)/(dx))+(logx)y=x^(-1/2logx)` is

Options

A. `y = x^(1-1/2log x)+cx^(-1/2 log x)`
B. `y . x^(1/2 log x) = x^(1/2 log x) + c`
C. `y = e^(((log x)^(2))/(2))(x+c)`
D. `y = e^(1/2(log x)^(2)) (x^(1-(1)/(2)(log x))-x^(-(1)/(2)log x))+ c`

Answer

A

Explanation

`(dy)/(dx)+(log x)/(x) y = x^(-1/2 log x)`
I.F `= e^(int(log x)/(x)dx)=e^((log x)^(2)/(2))=(e^(log x))^((log x)/(2))=x^(1/2 log x)`
`therefore" ""Solution is"x^(1/2 log x)*y=int dx`
`or" "yx^(1/2 log x)= x + c`