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Obtion the differential equation by elininating arbitrary constants A, B from the equation - `y=A cos (logx)+B sin (logx)`

Answer» `y=A cos (logx)+Bsin(logx)`
Differentiating w.r. to x, we get
`(dy)/(dx)=-A sin(logx)xx(1)/(x)+B cos (logx)xx(1)/(x)`
`x(dy)/(dx)=-Asin(logx)+B cos (logx)`
Again differentiating, we get
`x(d^(2)y)/(dx^(2))+(dy)/(dx)=-Acos(logx)xx(1)/(x)-B sin (log x)xx(1)/(x)`
`x^(2)(d^(2)y)/(dx^(2))+x(dy)/(dx)=-[A cos (logx)+B sin (logx)]`
`x^(2)(d^(2)y)/(dx^(2))+x(dy)/(dx)=-y`
`x^(2)(dy)/(dx)+y=0`


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