The solution of the differential equation `(1+y^2)+(x-e^(tan^-1y))dy/dx=0` is

Options

A. ` 2 x e^(tan^(-1)y)= e^(2tan^(-1)y)+C`
B. ` x e^(tan^(-1)y) = tan^(-1) y +C`
C. ` xe^(2tan^(-1)y)= e^(tan^(-1)y)+C`
D. ` (x-2) = C e ^(-tan^(-1)y)`

Answer

a

Explanation

Given differential equation is
` (1+y^(2))+(x-^(tan^(-1)y))`
`rArr (dx)/(dy)+ x/(1+y^(2)) = (e^(tan^(-1)y))/(1+y^(2))`
which is linear differential equation
Here, ` P = 1/(1+y^(2)),Q = (e^(tan^(-1)y))/(1+y^(2))`
` :. IF = e^(int P dy) = e^(int 1/(1+y^(2))dy)`
` = e^(tan^(-1))y`
` :. ` Solution is given by
` x * IF int Q * IF dy +C`
` xe^(tan^(-1)y) = int e^(tan-1y)/(1+y^(2))* e^(tan^(-1)y) +C/2 `
` rArrx e^(tan^(-1)y) = (e^(2tan^(-1)y))/y + C/2 `
` :. 2 x e^(tan^(-1)y)=e^(2tan^(-1)y)+C`