The solution of the differential equation `(x^(2)+4y^(2)-5)xd=(4x^(2)-3y^(2)-1)ydy`is

Options

A. `2/(sqrt(3))tan^(-1)(sqrt(3)((y^(2)-1)/(x^(2)-1)))-1/4 "In" (6((y^(2)-1)/(x^(2)-1))^(2)+2)-"In"sqrt(|x^(2)-1|)+c=0`
B. `2/(sqrt(3))tan^(-1)(sqrt(3)((y^(2)-1)/(x^(2)-1)))-1/4 "In" (6((y^(2)-1)/(x^(2)-1))^(2)+2)-"In"sqrt(|x^(2)-1|)+tanc=0`
C. `2/(sqrt(3))tan^(-1)(sqrt(3)((y^(2)-1)/(x^(2)-1)))-1/4 "In" (6((y^(2)-1)/(x^(2)-1))^(2)+2)-"In"sqrt(|x^(2)-1|)+e^(c)=0`
D. `2/(sqrt(3))tan^(-1)(sqrt(3)((y^(2)-1)/(x^(2)-1)))-1/4 "In" (6((x^(2)-1)/(y^(2)-1))^(2)+2)-"In"sqrt(|x^(2)-1|)+e^(c)=0`

Answer

A, B

Explanation

`(((x^(2)-1)+4(y^(2)-1))x)/((4(x^(2)-1)-3(y^(2)-1))y)=(dy)/(dx)`
Let `y^(2)-1=v(x^(2)-1)`
`implies (6v^(2)+2)/(4-3v)=((x^(2)-1)/x) (dv)/(dx)`
`=int (4-3v)/(6v^(2)+2)dv = int x/ (x^(2)-1) dx`
`implies2/(sqrt(3)) tan^(-1) (sqrt(3) ((y^(2)-1)/(x^(2)-1)))-1/4 "In" (6((y^(2)-1)/(x^(2)-1))^(2)+2)-"In" sqrt(|x^(2)-1|)+c=0`