The differential equation satistied by the family of curves `y=axcos(1/x+b)`, where a,b are parameters, is

Options

A. `x^(2)y_(2)+y=0`
B. `x^(4)y_(2)+y=0`
C. `xy_(2)-y=0`
D. `x^(4)y_(2)-y=0`

Answer

B

Explanation

Given `y=ax cos (1/x+b)`
`impliesy_(1)=-axsin (1/x+b)xx(-1/(x^(2)))`
`+cos (1/x+b)`
`impliesy_(1)=a/x sin (1/x+b)+a cos (1/x+b)`
`impliesxy_(1)=a sin (1/x+b)+y`
`impliesy_(1)+xy_(2)`
`=acos(1/x_h)(-1/(x^(2))+y_(1))`
`impliesx^(3)y_(2)=-a cos (1/x+b)`
`impliesx^(4)y_(2)+y=0`