यदि `2x=y^(1//3)+y^(-(1)/(3))` हो तो `(x^(2)-1)(d^(2)y)/(dx^(2))+x(dy)/(dx)` का मान होगा

Options

A. `9y`
B. `-9y`
C. `9`
D. `-9`

Answer

A

Explanation

दिया हैं
`2x=y^(1//3)+y^(-1//3)`

`2x=y^(1//3)+1/(y^(1//3))`

`2x=(y^(2//3)+1)/(y^(1//3))`

`2x*y^(1//3)=y^(2//3)+1`

`y^(2//3)-2x*y^(1//3)+1=0 `

चूंकि यह द्विघात समीकरण हैं अतः इसका मूल

`(y)^(1//3)=((xpmsqrt(x^(2)-1))^(3))/1`

`rArr" "y=(xpmsqrt(x^(2)-1))^(3)`

` :." "(dy)/(dx)`

`=3(xpmsqrt(x^(2)-1))^(2)``xx``(1pm(2x)/(2sqrt(x^(2)-1)))`

`=3(xpmsqrt(x^(2)-1))^(2)((sqrt(x^(2)-1)pmx)/(sqrt(x^(2)-1)))`

`=pm(3(xpmsqrt(x^(2)-1))^(3))/(sqrt(x^(2)-1))`

`=pm(3y)/(sqrt(x^(2)-1))`

` ((dy/dx))^(2)=(y_(1))^2=(9y^(2))/(x^(2)-1)`

`rArr(x^(2)-1)y_(1)^(2)-9y^(2)=0`

`x` के सापेक्ष अवकलन करने पर,

`rArr" "(x^(2)-1)2y_(1)y_(2)+2xy_(1)^(2)-18yy_(1)=0`

`rArr" "(x^(2)-1)y_(2)+xy_(1)=9y`

या `(x^(2)-1)(d^(2)y)/(dx^(2))+x(dy)/(dx)=9y`