Solve the differential equation: `(1+y^2) + ( x - e^(tan^-1 y) ) dy/dx = 0`

Options

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

Answer

C

Explanation

We have,
`(1+y^(2))+(x-e^(tan^(-1)))(dy)/(dx)=0`
`rArr" "(dx)/(dxy)+(x)/(1+y^(2))=(e^(tan^(-1)y))/(1+y^(2))" ...(i)"`
The linear differential equation with
`"I.F."=e^(int(1)/(1+y^(2))dy)=e^(tan^(-1)y)`
Multiplying (i) by I.F. and integrating, we get
`xe^(tan^(-1)y)=int(e^(2tan^(-1)y))/(1+y^(2))dy`
`rArr" "xe^(tan^(-1)y)=(1)/(2)inte^(2tan^(-1)y)d(2tan^(-1)y)`
`rArr" "xe^(tan^(-1)y)=(1)/(2)e^(2tan^(-1)y)+C`
`rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1))+K," where K"=2C`